1 Double Integrals over Rectangular Regions

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1 Contents ouble Integrals over Rectangular Regions ouble Integrals Over General Regions 7. Introduction Areas of General Regions Region Types ouble Integrals in Polar Coordinates 4 4 Triple Integrals 4. Rectangular Regions General Regions Triple Integrals in Cylindrical Coordinates 6 6 Triple Integrals in Spherical Coordinates 3 ouble Integrals over Rectangular Regions We now begin our study of multi-variable integration. In single variable calculus, we learned that the definite integral b a f(x) dx represents the area between the curve defined by f(x) and the x-axis, between x = a and x = b. This can be understood geometrically by thinking of f(x) dx as the area of a rectangle with an infinitesimally small base dx and a height of f(x). The operation of integration is to add up all of these areas to get the total area. So how does this idea extend to functions of two variables? A function of the form z = f(x, y) represents a surface in R 3. Let R be some rectangular region in the xy plane. valuating f over every point of R creates a solid pillar with a rectangular base and a curved top, where the shape of the top of the pillar is determined by the shape of the surface, as shown in the figure below. The volume of this solid can be calculated with multi-variable integration. Let R be the rectangular region in the xy plane such that a x b and c y d. We say that this region is the Cartesian product of the intervals [a, b] and [c, d] and we denote this as R = [a, b] [c, d] = {(x, y) a x b and c y d}. Note that this notation uses the symbol, but it is not at all related to the cross product of vectors. Let (x, y) be some point inside the rectangle R and let z = f(x, y) be a surface in R 3. Suppose we draw a small rectangle with dimensions x and y that is centered around (x, y). The area of this rectangle is A = x y. Then the volume V of the pillar that lies over the rectangle and under the surface z = f(x, y) is approximately the area of times the height of the surface at (x, y): V f(x, y) A = f(x, y) x y. Note that this is an approximation because the pillar is actually curved at the top since the function can change within the rectangle. However, in the limit as x, y, i.e., each dimension of becomes infinitesimal, the above approximation becomes exact: dv = f(x, y) da = f(x, y) dxdy,

2 Figure : Illustration of a solid under a surface and over the region [, ] [, 3]. which is the volume of the infinitesimal pillar centered at (x, y), with a dx-by-dy rectangular base. We obtain the total volume under the surface and over the region R = [a, b] [c, d] by adding up all of the differential volumes for the pillars centered at every point in R. In other words, we integrate f(x, y) over the region R. The total volume under the surface and over the region R is written symbolically as a double integral: V = R f(x, y) da. Since da = dx dy and R = [a, b] [c, d], we can write the above integral strictly in terms of x and y, which gives us an iterated integral: R f(x, y) da = d b c a f(x, y) dx dy. Notice that the double integral has an inner integral (with respect to x), and an outer integral (with respect to y). We interpret the double integral as follows: for a fixed value of y, we add up all of the pillar volumes along the x axis, which tells us the volume the slab made of all the pillars at that y value. This slab would have a volume W (y) = b a f(x, y) dx where the integration is done with respect to x, treating y as a constant. Then we add up all of the slab volumes for every point along the y axis to obtain the total volume of the solid: d d [ b ] V = W (y) dy = f(x, y) dx dy. c c In other words, we do the inner integral first, while treating the outer variable as constant. This results in a function of the outer variable only, and reduces the double integral to a single integral. Note that because the region we are integrating over is perfectly rectangular, we could have instead done this process with y as the inner variable and x as the outer a

3 variable. Therefore the integrals can be written in either order for this problem. This is known as Fubini s Theorem: d b c a f(x, y) dx dy = b d a c f(x, y) dy dx. However, as we will see in the next section, this is not the case when integrating over nonrectangular regions. Many of the common properties and intuitions from single integrals carry over to double integrals. For example, surfaces that lie below the xy plane contribute negative volume, just as curves that lie below the x-axis contribute negative area in single integrals. The linearity properties of single integrals apply to double integrals as well: c f(x, y) + c g(x, y) da = c f(x, y) da + c g(x, y) da for any constants c and c and any two-variable functions f and g. Another useful property to know is that if f(x, y) can be factored into a function of x only times a function of y only, i.e., f(x, y) = g(x)h(y), then b d a c f(x, y) dy dx = b d a c [ b g(x)h(y) dy dx = a ] [ d ] g(x) dx h(y) dy. c In other words, we can do the single integrals for g(x) and h(y) separately, then multiply them to the get the answer for the double integral. Functions that can be factored this way are said to be separable. xample.. Find the volume under the surface of f(x, y) = x y y + over the rectangle R = [, ] [, ]. Solution: We need to do the double integral f(x, y) da = x y y + dx dy = R [ ] x y y + dx We first do the inner integral with respect to x, treating y as constant and get x y y + dx = x= 3 x3 y xy + x = 3 y y + = 3 y +. Now plugging this into the outer integral we get y= 3 y + dy = 3 y + y = ( 43 ) + ( 3 ) = =. y= Now let us do the integral in the other order, to verify that it is the same: [ ] x y y + dy dx. The inner integral is x y y + dy = x y y + y y= y= x= dy. = (x + ) ( x ) = 3 x + 3.

4 Plugging this into the outer integral, we get 3 x + 3 dx = x3 + 3 x x= x= = 3 + =. xample.. Find the volume under the surface of f(x, y) = xe y R = [, 3] [, ]. over the rectangle Solution: Since this function can be written as f(x, y) = g(x)h(y) with g(x) = x and h(y) = e y, the double integral is the product of the two single integrals: 3 [ 3 ] [ ] f(x, y) da = xe y dy dx = x dx e y dy. The x integral is The y integral is R 3 Therefore the double integral is x dx = x=3 x = 9 = 5. xample.3. Calculate the iterated integral R x= y= e y dy = e y = e. y= f(x, y) da = 5 (e ). x + e x y dx dy. Solution: This function doesn t appear to be separable but if we split this into two integrals as e x x dx dy + dx dy y then each of these integrals has a separable integrand and can be reduced to the product of two single integrals: [ ] [ ] [ ] [ ] x dx dy + e x dx y dy [ ] [ ] [ ] [ ] = y + e x ln y x = ( )( ) + (e e )(ln() ln()) = + ( e ) ln(). xample.4. Calculate the iterated integral Solution: We first do the inner integral x + y dx dy. x + y dx.

5 To do so, make the substitution u = x + y. Since y is treated as constant, du = dx. We also need to transform the integral bounds. When x =, u = y and when x =, u = y +. We can then compute the inner integral as y+ u=y+ x + y dx = u du = 3 u3/ = ( (y + ) 3/ y 3/). 3 y We now to do the outer integral: ( (y + ) 3/ y 3/) dy = ( (y + )5/ ) y= 5 y5/ u=y = 4 5 ((5/ ) ( )) = 4 5 (5/ ). xample.5. Calculate the iterated integral 3 dy dx. (x + y) Solution: We first do the inner integral 3 (x + y) dy. We make the substitution u = x + y. Then du = dy. We also need to transform the integral bounds. When y =, u = x + and when y = 3, u = x + 3. We can then compute the inner integral as 3 (x + y) dy = We now to do the outer integral: x+3 x+ u du = u u=x+3 u=x+ x + x + 3 dx = ln(x + ) = [ln(5) ln(7) ln(3) + ln(5)] = ln ( 5 y= = x + x + 3. ln(x + 3) xample.6. Find the volume of the solid S enclosed by the paraboloid z = +x +(y ) and the planes z =, x =, x =, y = and y = 4. Solution: Notice that the 4 planes for x and y form the rectangle R = [, ] [, 4]. If we were to do the integral of z over R, this would give us the volume enclosed by the surface and the xy plane. However, we want the volume of the solid enclosed by the surface and the plane z =, so we would have to subtract the volume of the solid box B enclosed by z = and z =. Since R is a by 4 rectangle and the box has a height of, the volume of this box is V (B) = ()(4)() = 8. The volume of the solid S is thus V (S) = 4 ). + x + (y ) dy dx V (B). Computing the inner integral with respect to y, we get 4 + x + (y ) dy = y + x y + 3 (y )3 y=4 y= x= x= = (8 + 4x + 8/3) ( 8/3) = 4x

6 Now doing the outer integral, we get 4x dx = 4 3 x x x= x= = (4/3 + 4/3) ( 4/3 4/3) = We now subtract V (B) = 8 from this number to get the volume of the solid S: V (S) = = 64 3.

7 ouble Integrals Over General Regions. Introduction In the last section, we learned how to do the double integral of a function of two variables over a rectangular region. Now we will study how to integrate two variable functions over a general region in R. To demonstrate how this works, we will start with an example. xample: Integrate the function f(x, y) = 6xy over the triangular region with vertices (, ), (, ) and (, ). Solution: We first notice that the region can be described as the region enclosed by the curves y =, x = and y = x. One way to write this region is Figure : The triangular region. = { (x, y) x and y x}. In other words, we vary x between and for every fixed value of x, we vary y from to x. The iterated integral would therefore be written as 6xy da = x 6xy dy dx. Notice the order in which the integrals are written, with the x integral on the outside and the y integral on the inside. Unlike rectangular regions, we can not arbitrarily switch the order of double integrals when is a non-rectangular region. When doing any double integral, the outer integral must always have constant bounds but the bounds on the inner integral may be functions of the outer variable. oing the inner integral with respect to y gives us x Plugging this into the outer integral, we get y=x 6xy dy = xy 3 = x[(x) 3 3 ] = 6x 4. y= 6x 4 dx = 6 5 x= x= = 6 5. We could also have done this integral in the other order (with x as the inner variable) but it would require some modification since the outer integral bounds must be constants. The region can also be written as = { (x, y) y and y/ x }.

8 In other words, we vary y between and for every fixed value of y, we vary x from y/ to. The iterated integral in this case would be written as 6xy da = y/ 6xy dx dy. This shows that we can switch the order of integration for double integrals over general regions, so long as take care to describe the region in a way where the outer integral bounds are constant. oing the inner integral with respect to x gives us y/ x=y/ 6xy dx = 3x y = 3y [ ( y ) ] = 3y 3 4 y4. x= Plugging this into the outer integral, we get 3y 3 4 y4 dy = y 3 3 y5 y= y= which is the same answer we got the first time. = = = 6 5, xample.. escribe the region of integration and switch the order of integration for the iterated integral x f(x, y) dy dx. x Solution: The region being integrated over is = { (x, y) x and x y x}, which is the region bounded by the curves y = x and y = x. This region can also be written as = { (x, y) y and y x y}. So switching the order of integration results in x x f(x, y) dy dx = y y f(x, y) dx dy. Figure 3: The region bounded by y = x and y = x.

9 . Areas of General Regions In single variable calculus, we learned how to compute areas of regions between curves, but we can also do these problems using double integrals. Since da represents a infinitesimal area of the region, we can compute the area of by adding up all of these infinitesimal areas with an integral: da = A() where A() is the area of. Note that this is equivalent to finding the volume of the solid over whose height is f(x, y) = at every point in. xample.. Use a double integral to the find the area of the region bounded by g(x) = x and h(x) = x. Solution: We first need to find the x-values where the two curves intersect. Setting the two functions equal to each other, we get g(x) = h(x) x = x x = x = ±. Since x x when x, the region is written as so the area of the region is A() = = = { (x, y) x and x y x } da = x x dy dx = x dx = x 3 x3 x= x= y= x y dx = y=x ( x ) x dx = ( 3 ) ( + 3 ) = = 8 3. Figure 4: The region bounded by y = x and y = x. One application of double integrals is to use them to find the average value of a function over a general region. The average value f of (x, y) over is f = f(x, y) da A() where A() is the area of. Since A() = da,

10 we can write the average value as f = f(x, y) da da. xample.3. Find the average value of the function of the function f(x, y) = xy over the region bounded by x =, y = and y = x. Solution: We first compute the area of A() = x dy dx = We now compute the double integral The inner integral is Plugging this into the outer integral gives x dx = x x= 3 x3 = 3 = 3. x xy dy dx. x= y= xy dy = xy = x x(x ) = x x 5. x y=x Therefore the average value is.3 Region Types x x 5 dx = x x= 6 x6 = 6 = 3. f = A() x= f(x, y) da = /3 /3 =. As the past few examples have demonstrated, when computing a double integral over a general region, we always have to choose which order to do the integration. For some regions, like a triangular region, the double integral will be about the same difficulty no matter which order we do it in. However, for most regions, one order will make the integration easier than the other. We define a Type I region as one that is stated as = { (x, y) a x b and g(x) y h(x)} where h(x) g(x) for all x [a, b]. For regions of this type, a double integral should be done in the order b h(x) f(x, y) da = f(x, y) dy dx. a An example of a Type region is shown below. On the other hand, a Type II region is one that is stated as = { (x, y) c y d and g(y) x h(y)} where h(y) g(y) for all y [c, d]. For regions of this type, a double integral should be done in the order d h(y) f(x, y) da = f(x, y) dx dy. c g(x) g(y)

11 Figure 5: xample of a Type I region. Figure 6: xample of a Type II region. Some regions, like the triangular region from the first example, can be stated as either Type I or Type II regions. For such regions, the order of integration will not significantly change the difficulty of the double integral. More complicated regions may not fall into either category. For example, a diamond shaped region is neither Type I or Type II. When encountering these types of regions, the region must be split up into multiple Type I or Type II regions. xample.4. Integrate the function f(x, y) = xy over the region bounded by y = x, y = x and y =. A sketch of this region reveals that it is a Type II region. If we solve for x in both curves, we can state them as x = y and x = y. Setting these two equal to each other, we see that they intersect at y =, which is the largest y value in the region. Since x x for all x in the region, the region is written as and the double integral is = { (x, y) y and y x y } xy da = y y xy dx dy. oing the inner integral gives y xy dx = x= y y x y = x=y y[( y ) (y ) ] = y(4 4y + y 4 y 4 ) = y y 3. Plugging this into the outer integral gives y y 3 dy = y y= y4 = =. y=

12 Figure 7: The Type II region bounded by y = x, y = x and y =. What if we tried to do this problem using the other order of integration? This is still possible but would be more complicated since we would have to split the region into two Type regions, and, where = { (x, y) x and y x}, = { (x, y) x and y x}. Figure 8: Splitting a Type II region into two Type I regions. Since =, the integral can be computed as xy da = xy da + xy da = x xy dy dx + x xy dy dx. There are also situations where choosing the appropriate order of integration can actually make an integral easier. Consider the following example. xample.5. etermine the easiest order to do the integral y e xy da where is the region bounded by y = x, y = 4 and x =. Solution: This is a triangular region, which we can treat as either Type I or Type II. If we treat as a Type I region, the integral takes the form 4 4 y e xy da = y e xy dy dx. To do the inner y integral, we would have to do two integration by parts, which is doable but tedious. However, if we treat as a Type II region, the integral takes the form 4 y y e xy da = y e xy dx dy. The inner integral can now be done easily to get y y e xy dx = y x=y y exy = ye y y. x x=

13 We now split the outer integral into 4 ye y dy 4 y dy. The first integral can be done with the substitution u = y, du = y dy so 4 ye y dy = The second integral is so the double integral is 4 6 e u du = u=6 eu = (e6 ). u= y dy = y=4 y = 8 y= y e xy da = (e6 ) 8. xample.6. Find the value of the double integral using geometry 4 6 x dy dx. Solution: Since this integral is of the form da, it is just the area of the region. The curve y = 6 x represents the top half of a circle with radius 4. However, since x 4, the region being integrated over is the quarter of the circle that lies in the first quadrant. This would just be one fourth of the area of a circle with radius 4 so we can conclude that the integral is 4 6 x dy dx = 4 π(4) = 4π.

14 3 ouble Integrals in Polar Coordinates Certain types of regions in R are difficult to describe using Cartesian coordinates and would be more easily handled using polar coordinates. Recall that the polar coordinate system specifies the position of a point using a radial coordinate r and an angular coordinate θ. Given a point P (x, y), there is a unique line through P and the origin O. The angular coordinate of P is the angle this line makes with the +x-axis and the radial coordinate of P is the distance from O to P. The Cartesian coordinates of P can be stated in terms of the polar coordinates as x = r cos(θ), y = r sin(θ). Alternatively, we can state the polar coordinates in terms of the Cartesian coordinates as r = x + y, θ = tan ( y x The grid lines of polar coordinates correspond to curves where r or θ is constant, just like the horizontal/vertical grid lines of Cartesian coordinates correspond to constant x or y. A polar grid can be visualized as shown in the figure below. Throughout this section, ). Figure 9: Visualization of a polar grid. we will have to make use of some trigonometric identities, which are stated here for reference. Pythagorean Identities: sin (θ) + cos (θ) =, tan (θ) + = sec (θ), cot (θ) + = csc (θ) ouble-angle Identities: sin(θ) = sin(θ) cos(θ), cos(θ) = cos (θ) sin (θ) Half-Angle Identities: cos (θ) = [ + cos(θ)], sin (θ) = [ cos(θ)]

15 Polar coordinates can be used to simplify double integrals over regions that have circular symmetry. These types of regions are not rectangular in Cartesian coordinates but they are in polar coordinates. A polar rectangle is a region stated as = [a, b] [c, d] = {(r, θ) a r b, c θ d}. Some examples of polar rectangles, such as a disc, an annulus and a sector, are shown in the figure below below. In general, a polar rectangle is a sector of an annulus. In order to do double integrals in polar coordinates, we need to be able to state the Figure : Top Left: A disc in polar coordinates: = [, ] [, π]. Top Right: An annulus in polar coordinates: = [, ] [, π]. Bottom Left: A circular sector in polar coordinates: = [, ] [π/6, π/3]. Bottom Right: An annular sector: = [, ] [π/6, π/3]. differential area da in terms of the differentials dr and dθ. Consider a general polar rectangle = [r, r + r] [θ, θ + θ]. The area of this region can be found by computing the area of the outer sector and subtract the area of the inner sector. Recall that the area of a sector of a circle with radius r between the angles θ and θ is r (θ θ ). The area of the r by θ polar rectangle is thus A = (r + r) θ r θ = θ ( r + r r + r r ) = r r θ + θ r. In the limit as r and θ, the second term becomes negligible and we get the differential area is da = r drdθ.

16 It should make sense that the area of a polar rectangle is proportional to the radial coordinate r since they grow wider as r increases. Using this differential area formula, we can compute the double integral of a function over a region in polar coordinates as f(x, y) da = f(r cos θ, r sin θ) r dr dθ. If the region is a polar rectangle [a, b] [c, d], then this formula can be stated as f(x, y) da = d b c a f(r cos θ, r sin θ) r dr dθ. We can also use double integrals to find the area of any polar region as A() = da = r dr dθ. xample 3.. Integrate the function f(x, y) = x y over the top half of the disc centered at the origin with a radius of. Solution: The region of integration can be stated as the polar rectangle = [, ] [, π]. The integral is then π π x y da = (r cos θ) (r sin θ)r dr dθ = r 4 cos (θ) sin(θ) dr dθ. The integrand is a separable function of r and θ so the double integral is just the product of the two single integrals [ ] [ π ] r 4 dr cos (θ) sin(θ) dθ. The first integral is r 4 dr = 5 r5 r= r= = 3 5. The second integral can be done with the substitution u = cos(θ). Then du = sin(θ) dθ. The integral bounds will then be cos() = and cos(π) =, which gives us the integral π The double integral is thus cos (θ) sin(θ) dθ = x y da = = u du = u= 3 u3 = 3. xample 3.. Find the area of one of the leaves in the four-leaf clover given by the equation r = sin(θ). Solution: Upon drawing this region, we see that the leaf in the first quadrant intersects itself at the origin and comes in tangent to the origin at θ = and θ = π/. This region can then be described as u= = {(r, θ) r sin(θ), θ π/}.

17 Note that this region is not a polar rectangle since the radius is dependent on θ. The area is of the region is given by the double integral π/ sin(θ) A() = da = r dr dθ. The inner integral is r r=sin(θ) r= = sin (θ). Plugging this into the outer integral gives π/ sin (θ) dθ = π/ cos(4θ) dθ = [θ 4 ] θ=π/ 4 4 sin(4θ) = π 8. xample 3.3. Find the volume of the solid that lies inside both the cone z = x + y and the sphere x + y + z =. Solution: Since this region lies strictly above the xy plane, we can solve for z in the sphere equation to get z = x y. The volume of the solid between these surfaces can be found by computing the volume of the top surface minus the volume of the bottom surface. In this case the top surface is the sphere so the volume is computed with the integral V = x y da x + y da In polar coordinates, this integral becomes ( r r) r dr dθ = r r dr dθ θ= r dr dθ. To find the region, we first find the radius of the circle where the surfaces intersect. Setting the z coordinates of the surfaces equal to each other, we get r = r r = r r =. The region that we are integrating over is therefore a circle of radius /, centered at the origin. The volume is therefore V = π / r r dr dθ π / r dr dθ The first integral is computed with the substitution u = r. Then du = r dr and the integral bounds become u = and u = /. The inner integral then becomes π / u / du dθ = π u=/ 3 u3/ = π ( ) 3 = π 3/ 3 ( 3/ ). The second integral is easily done as π / The volume of the solid is thus u= r dr dθ = π r=/ 3 r3 r= = π 3 3/. V = π 3 ( 3/ ) π 3 3/ = π 3 ( 3/ ) = π 3 ( ).

18 xample 3.4. Find the volume of the ellipsoid x + y + z /9 =. Solution: Notice that the ellipsoid is symmetric with respect to the xy plane and we can therefore just find the volume of the top half and multiply it by. The height of the upper half can be found by solving for z and taking the positive square root x + y + z /9 = z = 9( x y ) z = 3 x y = 3 r. We now need to integrate this function over the region where the ellipsoid intersects the xy plane. By setting z = in the ellipsoid equation, we get that r = r = so the region of integration is the unit circle. Therefore the volume of the ellipsoid is π V = (3 r ) r dr dθ = π r r dr. Notice that the factor of π came from doing the θ integral immediately. The integrand for r was done in the last problem with the substitution u = r, which results in r r dr = u / du = u= 3 u3/ = 3 ( ) = 3. Therefore V = π 3 = 4π. Note that the volume for a general ellipsoid x /a + y /b + z /c = is known to be V = 4 πabc. In the case where a = b = c, the formula reduces to the volume of a sphere. For 3 this problem, we had a = b = and c = 3, which reduces to our answer 4π. xample 3.5. valuate the double integral in polar coordinates: x x u= x + y dy dx. Solution: First we must determine the region of integration, which is bound by x =, x =, y = and y = x x. Squaring each side of y = x x gives us which is stated in polar coordinates as y = x x x + y = x, r = r cos(θ) r = cos(θ). This is a circle of radius, which is centered at (, ). However, since the lower limit of integration for y is, we only integrate over the top half. This region is swept out by varying θ from to π/ and varying r from to cos(θ). We can therefore state the polar region as = {(r, θ) θ π/, r cos(θ). The integrand in polar coordinates is just x + y = r

19 so the double integral becomes π/ cos(θ) r r dr dθ = oing the inner integral with respect to r, we get cos(θ) Plugging this into the outer integral results in π/ cos(θ) r dr dθ. r dr = r= cos(θ) 3 r3 = 8 3 cos3 (θ). 8 3 π/ r= cos 3 (θ) dθ. Now we use the fact that cos (θ) = sin (θ) to write the integrand as The integral can therefore be written as cos 3 (θ) = cos (θ) cos(θ) = [ sin (θ)] cos(θ). 8 3 π/ [ sin (θ)] cos(θ) dθ. Now make the substitution u = sin(θ), which gives us du = cos(θ) dθ and the limits of integration become u = sin() = and u = sin(π/) =. The integral then becomes 8 u du = (u u= 3 u3 ) = 8 3 ( 3 ) = 6 9. xample 3.6. Find the area of the region above the line y = but below the circle x + y = 4. Solution: First we need to transform the expressions for these curves into polar coorindates. For the line y =, we get y = r sin θ = r = sin θ. Of course, the circle x + y = 4 is just stated as r = in polar coordinates. Now we need to to find the theta bounds for this region, i.e., the values of theta where the line intersects the circle. At the intersection, the radial coordinates must be equal so sin θ = sin θ = θ = π 6 or θ = π π 6 = 5π 6. These are the lower and upper bounds for the theta integral. The bounds for r would go from the line to the circle, i.e. r goes from to. The double integral for the area is thus oing the inner integral, we get / sin θ r dr = r r= sin θ 5π/6 π/6 / sin θ r=/ sin θ = u= r dr dθ. ( 4 ) sin = θ csc θ. Now recall that the anti-derivative of csc θ is cot θ, which allows us to the inner integral as 5π/6 π/6 csc θ dθ = θ + θ=5π/6 [ 5π cot θ = θ=π/6 3 ] [ π cot(5π/6) 3 + ] cot(π/6) = 5π 3 3 π 3 3 = 4π 3 3.

20 4 Triple Integrals Just as we can integrate functions of two variables with double integrals, we can also integrate functions of three variables with a triple integral. The region of integration for a triple integral would be a solid region in R 3. Triple integrals are therefore written as f(x, y, z) dv = f(x, y, z) dx dy dz where dv is the volume of a differential box with dimensions dx, dy and dz. For double integrals, we can interpret the value of the integral as being the volume of the solid that lies between a surface and the xy plane over some region. Triple integrals can not really be interpreted this way since they would represent the volume of a 4-dimensional object which is beyond our ability to visualize. However, there is another way we can physically interpret triple integrals. Consider a solid object with a non-uniform density that occupies the region. Suppose f(x, y, z) represents the density of the object at the point (x, y, z). Then dm = f(x, y, z) dv is the differential mass dm that occupies the differential box with volume dv around the point (x, y, z). The triple integral over would therefore give us the total mass M of the object: M = f(x, y, z) dv. In the case where f(x, y, z) =, i.e., the object has a uniform density of, this integral would tell use the volume of : V () = dv. Triple integrals are done just like double integrals: we start with the innermost integral and work our way outwards. Needless to say, all of the linearity properties of single and double integrals also apply to triple integrals. 4. Rectangular Regions If is a 3-dimensional rectangular region = [a, b] [c, d] [r, s], then a triple integral over can be written as the iterated integral f(x, y, z) dv = s d b r c a f(x, y, z) dx dy dz. Fubini s Theorem also applies to triple integrals over rectangular regions, although there are now 6 different, but equivalent, orders in which we could do the integral. The six possible ways we could arrange the differentials are dxdydz, dydzdx, dzdxdy, dzdydx, dydxdz, dxdzdy. If f(x, y, z) is separable in each variable, i.e., f(x, y, z) = g(x)h(y)p(z), then the triple integral is the product of three single integrals s d b [ s ] [ d ] [ b ] f(x, y, z) dx dy dz = p(z) dz h(y) dy g(x) dx. r c a r c a xample 4.. Integrate f(x, y, z) = x + y + 4z over the region = [, ] [, ] [, ] using the orderings dx dy dz and dz dx dy.

21 Solution: The integral for the first ordering is x + y + 4z dx dy dz. oing the innermost integral with respect to x gives x + y + 4z dx = x + xy + 4xz x= x= = + y + 4z. Now we plug this into the next inner integral with respect to y to get + y + 4z dy = y + y + 4yz y= y= Now we plug this into the outer integral for z to get 4 + 4z dz = 4z + z z= = ( z) ( + + 4z) = 4 + 4z. z= = (4 + ) ( 4 + ) = 8, which is the final answer. Now we ll do the integral with the other ordering: x + y + 4z dz dx dy. oing the innermost integral with respect to z gives x + y + 4z dz = xz + yz + z z= z= = (x + y + ) ( x y + ) = 4x + 4y. Now we plug this into the next inner integral with respect to x to get 4x + 4y dx = x + 4xy Now we plug this into the outer integral for y to get x= x= = + 4y. y= + 4y dy = y + y = (4 + 8) ( + ) = 8, which is the same answer as we got with the other ordering, as should be expected. y= xample 4.. Integrate f(x, y, z) = e 4x y+z over the region = [, ] [, ] [, ] using any ordering. Solution: First notice that this is a separable function: f(x, y, z) = e 4x y+z = e 4x e y e z so the triple integral is the product of three single integrals: [ ] [ ] [ e 4x y+z dx dy dz = e z dz e y dy ] e 4x dx.

22 The first integral is The second integral is z= e z dz = e z = e e. z= The third integral is e 4y dy = y= 4 e 4y = 4 (e 4 e 8 ). y= e x dx = x= ex = (e ). The triple integral is the product of these three single integrals: 4. General Regions x= e 4x y+z dx dy dz = 8 (e e )(e 4 e 8 )(e ). When setting up a triple integral over a general region, we must choose an ordering for the differentials that makes the integration as easy as possible. There are three types of common regions we will encounter. A Type I region is stated as = {(x, y, z) (x, y), u (x, y) z u (x, y)}. This represents the region between the two surfaces u (x, y) and u (x, y) and over the region in the xy plane. In this case, the triple integral would be done as [ ] u (x,y) f(x, y, z) dv = f(x, y, z) dz da. u (x,y) Notice that once the inner integral with respect to z is done, we would be left with a double integral over, which can be done with any of the double integral methods we already know. A Type II region is stated as = {(x, y, z) (y, z), u (y, z) x u (y, z)}. In this case, the triple integral would be done as [ ] u (y,z) f(x, y, z) dv = f(x, y, z) dx u (y,z) da. A Type III region is stated as = {(x, y, z) (x, z), u (x, z) y u (x, z)}. In this case, the triple integral would be done as [ ] u (x,z) f(x, y, z) dv = f(x, y, z) dy da. u (x,z)

23 xample 4.3. escribe the region of the solid in the first octant that lies under the plane z = 4 4x y as a Type I, Type II and Type III region. Then find the volume of by treating it as a Type I region. Note that this type of solid is called a tetrahedron. Solution: For each of the three types of regions, we will have to look at the projection of onto each coordinate plane. In each case, it would be a triangle with two vertices along the coordinate axes. If we treat as a Type I region, the innermost integral will be with respect to z, which goes from up to the plane z = 4 4x y. The outer two integrals will be a double integral over some region in the xy plane, which is bounded by the coordinate axes and the line obtained from setting z = in the plane equation, which is = 4 4x y y = x. This defines the region = {(x, y) x, y x}. The region can therefore be stated as a Type I region = {(x, y, z) x, y x, z 4 4x y}, which corresponds to the triple integral dv = x 4 4x y dz dy dx. Now we set the integral up as a Type II region. In this case, innermost integral would be with respect to x, which would go from x =, out to the plane x = (4 y z)/4, which we obtain be solving for x in the equation for the plane. The outer two integrals will be a double integral over some region in the yz plane, which is bounded by the coordinate axes and the line obtained from setting x = in the plane equation, which is z = 4 y. This defines the region = {(y, z) y, z 4 y}. The region can therefore be stated as a Type II region = {(x, y, z) y, z 4 y, x (4 y z)/4}, which corresponds to the triple integral dv = 4 y (4 y z)/4 dx dz dy. Now we set the integral up as a Type III region. In this case, innermost integral would be with respect to y, which would go from y =, out to the plane y = (4 4x z)/, which we obtain be solving for y in the equation for the plane. The outer two integrals will be a double integral over some region in the xz plane, which is bounded by the coordinate axes and the line obtained from setting y = in the plane equation, which is z = 4 4x. This defines the region = {(x, z) x, z 4 4x}. The region can therefore be stated as a Type III region = {(x, y, z) x, z 4 4x, y (4 4x z)/}, which corresponds to the triple integral dv = 4 4x (4 4x z)/ dy dz dx.

24 We have now set the integral up in 3 out of the 6 different ways possible. The other three possible orderings would be obtained by reversing the order of integration for the outer integrals in each of the 3 triple integrals we have already set up. We now find the volume of by doing the Type I triple integral x 4 4x y dv = dz dy dx. The innermost integral is 4 4x y dz = z z=4 4x y z= = 4 4x y. Plugging this into the next inner integral with respect to y gives us x y= x 4 4x y dy = 4y 4xy y = 4( x) 4x( x) ( x) y= = ( x)[4 4x ( x)] = ( x)( x) = 4 8x + 4x. Plugging this into the outer integral with respect to x gives 4 8x + 4x dx = 4x 4x + 4 x= 3 x3 = = 4 3. xample 4.4. Compute the triple integral where is bounded by the cylinder y + z = 9 and the planes x =, y = 3x and z = in the first octant. Solution: We will set this up as a Type I region. Solve for z in the cylinder equation and keeping in mind that z is positive in the first octant, gives us z = 9 y. Therefore z goes from to 9 y. Therefore the triple integral has the form [ ] 9 y z dz da where is the region in the xy plane bounded by the projections of the surfaces y = 3x, x = and y + z = 9. In the xy plane, z = so the projection of the cylinder is y = 9 y = 3. Therefore the region is the triangle bounded by x =, y = 3x and y = 3, which can be stated as = {(x, y) x, 3x y 3}. The triple integral can then be written as The innermost integral is 9 y z dv = 3 3x z dv z dz = z= 9 y z x= 9 y z dz dy dx. z= = (9 y ).

25 Plugging this into the next inner integral gives 3 3x (9 y ) dy = (9y 3 y3 ) 3 y=3x Plugging this into the outer integral gives us (8 7x + 9x3 ) dx = 7 (8x x x4 ) = [(7 9) (7x 9x3 )] = (8 7x + 9x3 ). x= x= = 7 ( ) = 7 4.

26 5 Triple Integrals in Cylindrical Coordinates The cylindrical coordinate system is a 3 coordinate system that is an extension of polar coordinates. A point P in cylindrical coordinates is specified by the ordered triple (r, θ, z) where r and θ are the polar coordinates of the projection of P onto the xy plane. In other words, we use polar coordinates for x and y and treat z as a Cartesian coordinate. The relations between the Cartesian coordinates and the cylindrical coordinates are thus x = r cos(θ), y = r sin(θ), z = z. Likewise, the inverse relationships can be used to state r, θ and z in terms of the Cartesian coordinates as r = ( y ) x + y, θ = tan, z = z. x xample 5.. Convert the point P = (r, θ, z) = (4, π, ) to Cartesian coordinates. Solution: Using the above formulas, we get x = 4 cos(π) = 4, y = 4 sin(π) =, z = so (x, y, z) = ( 4,, ) are the Cartesian coordinates of P. xample 5.. Convert the point P = (x, y, z) = (, 3, 5) to Cartesian coordinates. Solution: Using the above conversion formulas, r = ( ) = 3, θ = tan = π/, z = 5. so (r, θ, z) = (3, π/, 5) are the cylindrical coordinates of P. Cylindrical coordinates are ideally suited to do triple integrals over regions bounded by surfaces that have simple expressions when stated in cylindrical coordinates. Some examples of such surfaces are cylinders, cones and paraboloids. xample 5.3. State the equations for the following surfaces in cylindrical coordinates: a) the cylinder x + y = 4 b) The cone z = 3 x + y c) the paraboloid z = x + y +. Solution: a) Since x + y = r, the cylinder is stated in in cylindrical coordinates simply as r =. b) Since x + y = r, the cone is stated in in cylindrical coordinates as z = 3r. c) Since x + y = r, the paraboloid is stated in in cylindrical coordinates as z = r +. A differential volume in cylindrical coordinates is stated as dv = dx dy dz = r drdθdz. A triple integral of a function F (x, y, z) over a solid region is thus F (x, y, z) dv = F (r cos(θ), r sin(θ), z) r dz dr dθ.

27 Notice the order in which the differentials are written in the triple integral. ven though it is possible to use other orderings, most of the surfaces we will encounter will be of the form z = f(r), which means that z must be the inner integration variable. Notice that the cone and the paraboloid from the previous example can both be written as z = f(r). xample 5.4. Integrate the function f(x, y, z) = x + y + z over the cylinder bounded by the surfaces x + y = 9, z = and z = and z =. Solution: In cylindrical coordinates, the cylinder described in the problem represents a cylindrical rectangle, i.e., the iterated integrals will all have constant bounds. This region is stated in cylindrical coordinates as = {(r, θ, z) r 3, θ π z }. The integrand in cylindrical coordinates is stated as The triple integral is thus (x + y + z ) dv = f = x + y + z = r + z. π 3 (r + z ) r dz dr dθ = oing the innermost integral with respect to z gives r 3 + rz dz = r 3 z + 3 rz3 z= z= The next inner integral with respect to r is then 3 3r 3 + 3r dr = 3 4 r4 + 3 r r=3 The outer integral just adds a factor of π: π 3 = [r r] [ r3 3 r] = 3r3 + 3r. r= = = r 3 + rz dz dr dθ. π 97 4 dθ = π 97 4 = 97π. xample 5.5. Find the volume of the solid above the cone z = x + y and below the paraboloid z = 6 x y. Solution: The volume of this region in cylindrical coordinates is calculated with the triple integral V = dv = r dz dr dθ. The region is described by varying z from the bottom surface (the cone) to the top surface (the paraboloid), varying θ from to π and varying r from to some maximum radius R, where the two surfaces intersect. To find this maximum radius, we equate the z coordinates of each surface to find the r value where they intersect: r = 6 r r + r 6 = (r + 3)(r ) = r = 3,.

28 However, since r >, r = is the only acceptable solution. Therefore the triple integral for the volume of this region is V = π 6 r r r dz dr dθ. oing the innermost integral with respect to z gives 6 r z=6 r r dz = rz = r(6 r ) r = 6r r 3 r. r z=r The next inner integral with respect to r is then 6r r 3 r dr = 3r 4 r4 3 r3 r= The outer integral just adds a factor of π: π 6 3 r= dθ = π 6 3 = 3π 3. = = 6 3. xample 5.6. Integrate the function f(x, y, z) = xyz over the part of the cylinder x +y = 4 in the first octant and below the plane z = 4. Solution: The region described is the cylindrical rectangle The triple integral is thus xyz dv = = {(r, θ, z) r, θ π/, z 4}. π/ 4 (r cos(θ))(r sin(θ))z r dz dr dθ. Using the trig identity sin(θ) = sin(θ) cos(θ), this integral simplifies to π/ 4 r3 z sin(θ) dz dr dθ. Since the bounds of integration are constant and the integrand is separable, the triple integral is the product of the three single integrals: [ 4 ] [ ] [ ] π/ z dz r 3 dr sin(θ) dθ. The z integral is The r integral is The θ integral is π/ 4 z dz = z=4 z = 8. z= r 3 dr = r= 4 r4 = 4. r= sin(θ) dθ = θ=π/ 4 cos(θ) = 4 ( ) =. The value of the triple integral is the product of these three numbers: [ 4 ] [ ] [ ] π/ z dz r 3 dr sin(θ) dθ = (8)(4)(/) = 6. θ=

29 xample 5.7. Find the volume of the solid that inside both the sphere x + y + z = 9 and the cylinder x + y =. Solution: By symmetry, we can just find the volume of the top half of this solid (the portion above the xy plane), and multiply it by. The volume of this region in cylindrical coordinates is calculated with the general triple integral V = dv = r dz dr dθ. The region is described by varying z from the xy plane to the top of the sphere, varying θ from to π and varying r from to. In cylindrical coordinates, the equation of the sphere is r + z = so z = 9 r gives the height for the top half of the sphere. The volume integral is therefore V/ = dv = π 9 r oing the innermost integral with respect to z gives 9 r r dz = rz z= 9 r z= The next inner integral with respect to r is then r 9 r dr r dz dr dθ. = r 9 r. which is done with the substitution u = 9 r. Then du = r dr and the bounds become u = 9 and u = 8. The integral then becomes 8 The outer integral just adds a factor of π: 9 u / du = 3 (93/ 8 3/ ) = 3 (7 83/ ). so the total volume is V/ = π 3 (7 83/ ) dθ = π 3 (7 83/ ) V = 4π 3 (7 83/ ).

30 6 Triple Integrals in Spherical Coordinates The spherical coordinate system is another 3 coordinate system that is especially useful for doing triple integrals over regions with spherical symmetry. Let P = (x, y, z) be a point in R 3. The spherical coordinates (ρ, θ, φ) of P are defined as follows. Consider the line segment L that connects O to P. The radial coordinate ρ is the length of this line segment, i.e., the distance from O to P. The longitudinal coordinate θ is the angle that the projection of L onto the xy plane makes with the +x-axis. The latitudinal coordinate φ is the angle the L makes with the +z-axis. The values of ρ must always be greater than or equal to zero, since it represents a distance, with ρ = representing the origin. The values of θ are always between and π. However, the values of φ are restricted to be between and π to avoid any ambiguity in specifying a point. In other words, φ always represents the acute angle between the line segment OP anf the +z-axis. Note that the geographic coordinate system of longitude and latitude, commonly used by pilots and sailors for navigation, is a special case of spherical coordinates for which ρ is constant (the radius of earth). This gives a two-dimensional coordinate systems (θ, φ) where the θ coordinate specifies the longitude (the east/west position relative to the prime meridian) and the φ coordinate specifies the latitude (the north/west position relative to the equator). Note that every point on the equator has a φ coordinate of π/ and the north/south poles have φ coordinates of and π, respectively. The relations between the spherical coordinates and the Cartesian coordinates are x = ρ sin(φ) cos(θ), y = ρ sin(φ) sin(θ), z = ρ cos(φ). We can state the radial coordinate ρ in terms of the Cartesian coordinates as ρ = x + y + z. It is also possible to state θ and φ in terms of x, y, z but these relationships will never be needed for us. xample 6.. Find the Cartesian coordinates of the point (ρ, θ, φ) = (, π/6, π/4). Solution: Using the spherical coordinate conversion formulas, we have x = ρ sin(φ) cos(θ) = sin(π/6) cos(π/4) = =, y = ρ sin(φ) sin(θ) = sin(π/4) sin(π/6) = z = ρ cos(φ) = cos(π/4) = =. 3 = 3, xample 6.. escribe the following surfaces given in spherical coordinates: a) ρ =, b) θ =, c) φ = π/4, d) ρ = sin(θ) sin(φ). Solution: a) Setting ρ = x + y + z =, we get that the surface has the form x + y + z = 4, which is the equation of a sphere with radius, centered at the origin. In general, any surface of the form ρ = c is a sphere of radius c centered at the origin.

31 b) Setting θ =, we get x = ρ sin(φ), y =, z = ρ cos(φ) for ρ and φ π. For any fixed value of ρ, these are the parametric equations for a half-disc in the xz plane on the side where x. By varying ρ over every possible value from to infinity, these curves would sweep out the entire half of the xz plane where x. In general, a surface of the form θ = c represents a half-plane. c) Setting φ = π/4, we get We can eliminate θ since Since z = ρ /, we can eliminate ρ to get x = ρ cos(θ), y = ρ sin(θ), z = ρ. x + y = ρ (cos (θ) + sin (θ)) = ρ. x + y = z z = ± x + y. This surface represents two cones, one pointing upward from the origin and the other pointing downward. However, since z = ρ, we only the get the upward pointing cone z = x + y. In general, any surface of the form φ = c is a cone with its apex at the origin. It will be opening upwards if < φ < π/ and downwards if π/ < φ < π. The surface φ = π/ is just the xy plane, since every line segment in the xy plane is orthogonal to the +z-axis. d) Notice that since y = ρ sin(φ) sin(θ), we can eliminate θ and φ since Now we eliminate ρ to get ρ = y ρ ρ = y. x + y + z = y x + (y /) /4 + z = x + (y /) + z = /4, which is the equation for a sphere of radius /, centered at (, /, ). The previous example demonstrates that spherical coordinates are good for describing regions involving spheres and cones and are therefore ideal for doing triple integrals over regions with these shapes. The differential volume dv is more complicated in spherical coordinates and turns out to be dv = dx dy dz = ρ sin(φ) dρ dφ dθ. This volume can be interpreted as the area of a differential patch of a sphere da = dθdφ bounded by two lines of latitude and two lines of longitude, times a differential radius dρ. The ρ factor in the differential volume arises because the patches grow wider as ρ increases. The sin(φ) factor in the differential volume has a special significance. Notice that sin(φ) = when φ = and φ = π, which correspond to the north and south poles. Since all lines of longitude converge at the poles, the patches become infinitely thin there and therefore have

32 zero area. The maximum value of sin(φ) is, which occurs atφ = π/, which corresponds to points on the equator. Therefore the differential volume is largest for points the equator, which is where the lines of longitude have the greatest spacing between them. With the differential volume above, we can state the triple integral of a function f(x, y, z) over a region in spherical coordinates as f(x, y, z) dv = f(ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ sin φ dρ dφ dθ. Notice the order in which the differentials are written. All triple integrals are done in this order since θ and φ will always have constant bounds. In almost all problems, ρ will also have constant bounds as well. However, it is not unheard of for ρ to be a function of θ or φ or both, but these problems are very rare. = xample 6.3. erive the volume of a sphere of radius R. Solution: In spherical coordinates, the solid ball B of radius R is the spherical rectangle B = {(ρ, θ, φ) ρ R, θ π, φ π}. Therefore its volume is V = B dv = π π R ρ sin φ dρ dφ dθ. Since all bounds are constant and the integrand is separable, this reduces to the product of three integrals: π π R [ π] [ ] R dθ sin φ dφ ρ dρ = [π] cos φ 3 ρ3 = π 3 R3 = 4 3 πr3. xample 6.4. erive the volume of the solid region that inside both the code φ = π/6 and the sphere x + y + z =. Solution: In spherical coordinates, this region is the spherical rectangle Therefore its volume is V = = {(ρ, θ, φ) ρ, θ π, φ π/6}. B dv = π π/6 ρ sin φ dρ dφ dθ. Since all bounds are constant and the integrand is separable, this reduces to the product of three integrals: π π/6 [ ] [ π/6 ] ( ) dθ sin φ dφ ρ dρ = [π] cos φ 3 ρ3 3 = π 3 = π 3 ( 3). xample 6.5. Integrate f(x, y, z) = (x + y + z ) 3/ over the portion of the spherical shell ρ in the first octant. xample: First notice that we state the integrand as f(x, y, z) = (x + y + z ) 3/ = (ρ ) 3/ = ρ 3.

33 The region described is the spherical rectangle The triple integral is thus = {(ρ, θ, φ) ρ, θ π/, φ π/}. f(x, y, z) dv = π/ π/ ρ 3 ρ sin φ dρ dφ dθ. Since all bounds are constant and the integrand is separable, this reduces to the product of three integrals: π/ π/ [ dθ sin φ dφ ρ dρ = π ] [ π/ ] cos φ ln(ρ) = π ( ( ))(ln() ln()) = π ln(). xample 6.6. Convert the following integral to spherical coordinates: x x y x +y xy dz dy dx. Solution: The bounds on z go from the cone z = x + y up to z = x y, which is the top half of the the unit sphere. The xy projection of the region is bounded by x =, x =, y = and y = x, which is the quarter of the unit circle in the first quadrant. Therefore the region of integration is the region between the cone z = x + y and the unit sphere x + y + z = that lies in the first octant. The equation for the unit sphere is just ρ =. To find the equation of the cone, we set z = x + y = ρ sin φ cos θ + ρ sin φ sin θ = ρ sin φ(cos θ + sin θ = ρ sin φ). Since z = ρ cos φ in spherical coordinates, we get z = ρ cos φ = ρ sin φ cos φ = sin φ φ = π 4 so φ = π/4 is the equation of the cone. The region of integration is thus = {(ρ, θ, φ) ρ, θ π/, φ π/4} and the triple integral is xy dv = π/ π/4 (ρ sin φ cos θ)(ρ sin φ sin θ)ρ sin φ dρ dφ dθ. Since all bounds are constant and the integrand is separable, this reduces to the product of three integrals: π/ cos θ sin θ dθ π/4 sin 3 φ dφ ρ 4 dρ.