Chapter 14 Multiple Integrals..1 Double Integrals, Iterated Integrals, Cross-sections.2 Double Integrals over more general regions, Definition, Evaluation of Double Integrals, Properties of Double Integrals.3 Area and Volume by Double Integration, Volume by Iterated Integrals, Volume between Two surfaces.4 Double Integrals in Polar Coordinates, More general Regions.5 Applications of Double Integrals, Volume and First Theorem of Pappus, Surface Area and Second Theorem of Pappus, Moments of Inertia.6 Triple Integrals, Iterated Triple Integrals.7 Integration in Cylindrical and Spherical Coordinates..8 Surface Area, Surface Area of Parametric Surfaces, Surfaces Area in Cylindrical Coordinates.9 Change of Variables in Multiple Integrals, Jacobian
Chapter 14 Multiple Integrals. 14.5 Triple Integral, Upper and Lower Limits 14.5 Projections onto coordinate planes. 14.5 Mass, Moments, Centroid, Moment of Inertia.
Let B = [a, b] [c, d] [r, s] be a rectangular solid, and f (x, y, z) be a continuous scalar function defined on B. Divide B into l m n smaller rectangular solids. Label each small rectangular solid by C ijk, where 1 i l, 1 j m and 1 k n. Inside each such C ijk, pick a point P ijk = (x ijk, y ijk, z ijk ). Denote the volume of C ijk by V. Then we may form the Riemann sum: l m n i=1 j=1 k=1 f (P ijk ) V.. Definition. The triple integral of f over the rectangular box B is defined l m to f (x, y, z) dv = lim f (Pijk ) V.. B n l,m,n i=1 j=1 k=1 Remark. The triple integral over non-rectangular region can be defined similarly as that of double integral by summing up the smaller rectangles inside the region in the Riemann sum, and then obtain the limit as triple integral by using the Riemann sum.
Remarks. In general, it is difficult to compute the triple integral of scalar function over a solid region in space. So one can use the same idea which work very well in double integral, to evaluate the triple integral via iterated integral. Of course, we need to more. understanding of the region first.
Fubini s Theorem for non-rectangular region. Let D be a solid region in space.if D = { (x, y, z) R 3 (x, y) R, and z min (x, y) z z max (x, y) for all (x, y) R }, where z max (x, y) and z min (x, y) are continuous functions defined in the region D in xy-plane. Let f (x, y, z) be a scalar function defined in the region D, ( zmax ) (x,y) then f (x, y, z) dv = f (x, y, z)dz da.. D R z min (x,y) Remark. (1). The region R in xy-plane is in fact the shadow of the D under the projection from R 3 onto xy-plane, or by simply forgetting the z-coordinate. The idea is to allow the point P(x, y, ) varies within the region D, and then draw a line through P perpendicular to xy-plane, which will enter the solid D when z reaches z min (x, y) and then exit the solid D when z reaches z max (x, y). (2). The triple integral has similar properties like the double integral, here we con t repeat in stating these properties. (3). It is not necessary to project the solid D onto xy-plane, one can project D onto xz-plane, or yz-plane, in these cases, we should use function y = y(x, z) or x = x(y, z) respectively.
Sketch. Sketch the region D along with its "shadow" R (vertical projection) in the xy-plane. Label the upper and lower bounding surfaces. of D and the upper and lower bounding curves of R.
Find the z-limits of integration. Draw a line M passing through a typical point (x, y) in the shadow R parallel to the z-axis. As z increases, M enters the solid region D at z = z min (x, y) = f 1 (x, y) and. leaves at z = z max (x, y) = f 2 (x, y). These are the z-limits of integration
Find the y-limits of integration. Draw a line L through (x, y) parallel to the y-axis. As y increases, L enters R at y = y min = g 1 (x) and leaves at. y = y max (x) = g 2 (x). These are the y-limits of integration.
Find the x-limits of integration. Choose x-limits that include all lines through R parallel to the y-axis (x = a and x = b in the preceding figure). These are the x-limits of integration. The integral is x=b y=g2 (x) z=f1 (x,y). x=a y=g 1 (x) z=f 2 (x,y) F(x, y, z) dz dy dx.. Follow similar procedures if you change the order of integration. The "shadow" of region D lies in the plane of the last two variables with. respect to which the iterated integration takes place.
Example. Evaluate the triple integral dxdydz I =, where the solid D (1 + x + y + z) 3 D is bounded by the planes x. + y + z = 1, x =, y =, z =. Solution. Let R = { (x, y ) x + y 1, x, y } be the projection image of D onto xy-plane. Then D = { (x, y, z) x 1, y 1 x, z 1 x y }. ( dxdydz 1 x y ) I = D (1 + x + y + z) R 3 = dz (1 + x + y + z) 3 dxdy [ ] 1 1 x y = R 2(1 + x + y + z) 2 dxdy z= = 1 [ 1 2 R (1 + x + y) 2 1 ] dxdy 4 = 1 1 1 y dx 2 (1 + x + y) 2 dy 1 16 = = 1 2 ln 2 5 16.
Example. Evaluate D x 2 + z 2 dv, where D is the solid region bounded by the paraboloid S : y = x 2 + z 2 and the plane. π : y = 4. Solution. First determine the intersection of these surfaces E and S, take any intersection point P(x, y, z), so x 2 + z 2 = y = 4, so the intersection is a circle C : { (x, 4, z) x 2 + z 2 = 2 2 } in the plane π. We project D onto xz-plane, then its shadow R is bounded the image C of C. For any point Q(x, y, z) in D, then (x,, z) is in R, so we have y bottom (x, z) = x 2 + z 2 2 2 = 4 = y top (x, z), this means that the plane π lies above the paraboloid S over the D, the solid D = { (x, y, z) x 2 + z 2 4, x 2 + z 2 y 4 }. So ymin D x 2 + z 2 (x,z)=4 dv = x 2 + z 2 dy da xz = D y min (x,z)=x 2 +z 2 (4 x 2 z 2 ) 2π 2 x 2 + z 2 da xz = (4 r 2 ) r 2 rdrdθ = D 2 [ 4r 2π 4r 2 r 4 3 ] 2 dr = 2π 3 r5 = 128π 5 15.
Example. Determine the volume of the solid D bounded above by the. plane z = y + 2 and bounded below by the paraboloid z = x 2 + y 2. Solution. Let P(x, y, z) be any point in the intersection of the plane and the paraboloid, we have x 2 + y 2 = z = x + 2, i.e. x 2 + (y 1 2 )2 = 9 4. Then shadow R of D on xy-plane is given by { (x, y, ) x 2 + (y 1 2 )2 = 9 4 }. It follows that the volume of D is given by dv = (z top z bottom ) dzda D R = (y + 2 x 2 y 2 ) da. R Remark. We will complete the calculation in the following example.
Example. Determine the volume of the solid D bounded above by the. plane z = y + 2 and bounded below by the paraboloid z = x 2 + y 2. Solution. If we project the solid D onto yz-plane, then its shadow T in yz-plane is bounded by z = y 2 and the line z = y + 2. In this case, in yz-plane, we have T = { (y, z) 1 y 2, y 2 z y + 2 }. For any point P(, y, z) T, the line through P(, y, z), parallel to x-axis meet the paraboloid z = x 2 + y 2 at Q(x max, y, z) and Q (x max, y, z), so x max (y, z) = z y 2, and x min (y, z) = z y 2. The volume of D is xmax 2 y+2 z y 2 given by dv = dx da zy = dx dz dy D T x min 1 y 2 z y2 2 y+2 2 [ 2 = 2 z y 2 dzdy = 2 (z y 2) ] 3/2 y+2 dy 1 y 2 1 3 y 2 = 4 2 + y y 3 1(2 2 ) 3/2 dy = 4 3/2 ( ) 9 3/2 3 3/2 4 u2 du = 27 π/2 cos 4 θ dθ = 81 4 π/2 32 π. Remarks. In (*) we use u = y 1 2, and in ( ), we use u = 3 2 sin θ.
Example. Find the volume of the solid region D enclosed by the surfaces. z = x 2 + 3y 2 and z = 8 x 2 y 2. Solution. To find the limits of integration for evaluating the integral, we first sketch the region D. The surfaces (as shown above) intersect on the elliptical cylinder x 2 + 3y 2 = 8 x 2 y 2 or x 2 + 2y 2 = 4, z >. The boundary of the shadow R of D, after the projection onto xy-plane, is an ellipse with the same equation: x 2 + 2y 2 = 4. The upper boundary of R is the curve y = (4 x 2 )/2. The lower boundary is the curve y = (4 x 2 )/2. In this case, one can see that 2 x 2. It remains to find the z-limits of integration. The line M passing through a typical point (x, y) in R parallel to the z-axis enters D at z = x 2 + 3y 2 and leaves at z = 8 x 2 y 2.
Example. Find the volume of the solid region D enclosed by the surfaces. z = x 2 + 3y 2 and z = 8 x 2 y 2. Solution. The volume of the solid D is 2 ymax = (4 x 2 )/2 dv = = = = = D 2 y min = (4 x 2 )/2 2 (4 x 2 )/2 2 2 2 2 2 2 2 = 4 2 3 (4 x 2 )/2 (8 2x 2 4y 2 ) dy dx [(8 2x 2 )y 43 y3 ] y= (4 x 2 )/2 y= (4 x 2 )/2 zmax =8 x 2 y 2 z min =x 2 +3y 2 dx ( 4 x 2(8 2x 2 2 ) 8 ( 4 x 2 ) 3/2 ) dx 2 3 2 [ ( 4 x 2 ) 3/2 8 8 ( 4 x 2 ) 3/2 ] dx 2 3 2 2 2 dz dy dx (4 x 2 ) 3/2 dx = 8π 2. (* after plugging x = 2 sin u.)
Example. Set up the limits of integration in dydzdx, for evaluating the triple integral of a function F(x, y, z) over the tetrahedron D with vertices. (,, ), (1, 1, ), (, 1, ), and (, 1, 1). Solution. We sketch D along with its "shadow" R in the xz-plane. The upper (right-hand) bounding surface of D lies in the plane y = 1. The lower (left-hand) bounding surface lies in the plane y = x + z. The upper boundary of R is the line z = 1 x. The lower boundary is the line z =. R = { (x, z) x 1, z 1 x }. It remains to find the y-limits of integration. A line through a typical point (x, z) in R parallel to the y- axis enters D at y = x + z and leaves at y = 1. The required integral is 1 1 x 1 F(x, y, z) dv = F(x, y, z) dy dz dx. D x+z
Example. Set up the limits of integration in dz dy, dx, for evaluating the triple integral of a function F(x, y, z) over the tetrahedron D with vertices. (,, ), (1, 1, ), (, 1, ), and (, 1, 1). The required integral is 1 F(x, y, z) dv = D 1 y x x F(x, y, z) dz dy dx.
Example. Determine the volume of the solid region D in the first octant bounded by the coordinate planes and the surface z. = 4 x 2 y Solution. The level surface z = 4 x 2 y intersects the xy-plane at a curve C : 4 = x 2 + y. It follows that the shadow R of D on the xy-plane is given by R = { (x, y) x 2, y 4 x 2 }. In this case, the top is given by z max (x, y) = 4 x 2 y. Then the volume of D is given by (4 x 2 y) da = = = 1 2 2 4 x 2 2 2 R 2 ] 4 x 2 (4 x 2 y) dy dz = [(4 x 2 )y y2 2 [(4 x 2 ) 2 (4 x2 ) 2 ] dx = 1 2 (4 x 2 ) 2 dx 2 2 (16 8x 2 + x 4 ) dx = 1 ( 2 16 2 8 3 23 + 25 5 = 16 32 3 + 16 5 = 16 (1 2 3 + 1 5 ) = 128 15. )
Example. Determine the volume of the solid region D in the first octant bounded by the coordinate planes, the plane x + y = 4, and the. cylinder y 2 + 4z 2 = 16. Solution. The shadow R of D on the yz-plane is given by R = { (y, z) x, y, and y 2 + 4z 2 16 } = { (y, z) z 2, y 2 4 z 2 }. In this case, the top is given by x max (y, z) = 4 y. Then the volume of D is given by 2 2 4 z 2 (4 y) da = (4 y) dy dz = = R 2 (8 4 z 2 2(4 z 2 ) ) dz 2 = 8 1 4 π 22 8 2 + 2 3 23 = 8π 32 3. ] 2 4 z 2 [4y y2 dz 2
Example. Given D dv = 1 as. an equivalent iterated integral in the order (a) dy dz dx 1 1 1 y x 2 dzdydx, rewrite the integral Solution. For any point P(x,, ), the line through P parallel to y-axis meet the solid D at the curve y = x 2 at Q, so Q(x, x 2, ). Then the line through point Q parallel to z-axis meets the top y + z = 1 before leaving the solid D at R. Hence R(x, x 2, 1 x 2 ). The shadow S of D on the xz-plane is S = { (x,, z) 1 x 1, z 1 x 2 }. (Continue..)
Example. Given D dv = 1 as. an equivalent iterated integral in the order (a) dy dz dx 1 1 1 y x 2 dzdydx, rewrite the integral Solution. For any point P(x, z, ) in S, the line through P first passes through the cylinder (side) y = x 2 before entering the solid D, and then meets the plane (top) y + z = 1 before leaving the solid D, it follows that y min = x 2, and y max = 1 z. Hence, we have 1 1 x 2 1 z dv = dy dz dx. D 1 x 2
Example. The figure shows the region of integration for the integral 1 1 1 y x f (x, y, z) dz dy dx. Rewrite this integral as an equivalent iterated. integral in dy dz dx. Solution. For any point A(x,, ) on the x-axis, the line through A parallel to y-axis meets the cylinder y = x at a point B(x, x, ). Then the line through B parallel to z-axis meets the plane z = 1 y at C(x, x, 1 x). It follows that the shadow R of solid D onto xz-plane is given by R = { (x, z) x 1, z 1 x }. For any point P(x,, z) in the shadow, the line through P parallel to y-axis meets the cylinder y = x at the point (x, x, z) before entering the solid D and and meets the plane z = 1 y at the point (x, 1 z, z), it follows that y min = x, and y max = 1 z. It follows that D = { (x, y, z) x 1, z 1 x, x y 1 z }. Then we have 1 1 x 1 y f (x, y, z) dz dy dx = 1 1 x 1 z x dy dz dx.
Example. The figure shows the solid region D of integration for the integral 1 1 x 2 1 x f (x, y, z) dy dz dx. Rewrite this integral as an. equivalent iterated integral in the dz dy dx order. Solution. The shadow R of D onto xy-plane is the triangle given by R = { (x, y) x 1, y 1 x }. Note that the top of the solid D is the graph z = 1 x 2, and the bottom of D is xy-plane, i.e. z max (x, y) = 1 x 2, and z min (x, y) =. The desired iterated integral is 1 1 x 1 x 2 f (x, y, z) dz dy dx.
Example. The figure shows the solid region D of integration for the integral 1 1 x 2 1 x f (x, y, z) dy dz dx. Rewrite this integral as an. equivalent iterated integral in the dz dy dx order. Solution. For any P(x, y, z) in the intersection curve C (in red) of the two surfaces z = 1 x 2, and y = 1 x. Then P(x, y, z) satisfies the equations: z = 1 x 2, and y = 1 x. It follows that P(x, 1 x, 1 x 2 ). After projecting the point P(x, 1 x, 1 x 2 ) of the curve C onto yz-plane, we have Q(, 1 x, 1 x 2 ), set y = 1 x, and z = 1 x 2, i.e. z = (1 x)(1 + x) = (1 x)( 1 x) = y( 2 + y) = 2y y 2 for all y 1. Let R 1 = { (y, z) y 1, z y(2 + y) } and R 2 = { (y, z) y 1, y(2 + y) z 1 }. It follows that the integral is given by 1 2y y 2 1 z dx dz dy + 1 1 2y y 2 1 y dx dz dy.