Ma MULTIPLE INTEGRATION

Transcription

1 Ma 7 - MULTIPLE INTEGATION emark: The concept of a function of one variable in which y gx may be extended to two or more variables. If z is uniquely determined by values of the variables x and y, thenwesayz is a function of x and y, and write z fx,y. Thus for each pair of values x and y in the domain of f, fx,y gives one value of z. Double Integrals ecall that if fx then a b fxdx represents the area under f between x a and x b. y y=f(x) a b x Now consider a function fx,y of two variables x and y. ThenI fx,yda denotes the double integral over the region of the function fx,y. Actually when f is positive I is the volume under f which is enclosed by f, its projection onto the x,y plane and the shell of the projection. If we imagine a grid in the x,y plane then ΔA ΔxΔy ΔyΔxand da dxdy dydx fx,ydydx. I fx,ydxdy

2 If we are given the boundaries of in terms of y as a function of x, i.e. Then b fx,yda a f x fx,ydy dx. f x On the other hand if we are given the boundaries of in a form in which x is a function of y as indicated below

3 Then d fx,yda c g y fx,ydx dy g y Evaluate x y 3 da where is the region contained by the lines x,x,y,and y 3. x y 3 3 da x y 3 3 dxdy y3 x 3 dx dy y 3 x 3 dy 3 3 y dy y 3 dy 7 3 y We may also calculate the double integral by integrating with respect to y first 3

4 x y 3 3 da x y 3 3 dydx x y 3 dy dx x y x dx dx x Thus x y 3 dydx x y 3 dxdy. This is true in general. However, one must make sure that the limits of integration are correct. Evaluation of Double Integrals Here are a couple of examples of how one evaluates more complicated double integrals. Evaluate ln8 lny e xy dxdy ln8 lny e xy dxdy ln8 lny e x dx e y dy ln8 e x lny e y dy ln8 e y y dy ln8 ye y dy ln8 e y dy We use integration by parts to evaluate ln8 ye y dy with u y and dv e y dy ln8 lny e xy dxdy ye y ln8 ln8 e y dy ln8 e y dy 8ln8 e e y ln8 8ln8 e 6 e 8ln8 e 6. Evaluate x y 3 da where is the triangle with vertices at,,,,,. Solution: The triangle is shown below We will set up the integration in two ways. Consider first y x 4

5 x y 3 dxdy Taking a horizontal strip parallel to the x axis we see that x goes from the y axis to the line x y, whereas y goes from to. Thus x y 3 dxdy y x y 3 dxdy If we now consider x y 3 dydx then using a vertical strip parallel to the y axis we see that y goes from the line y x to. so we have x y 3 dydx x x y 3 dydx Find the volume of the solid whose base is in the x,y plane and is the triangle bounded by the x axis, the line y x and the line x, while the top of the solid is the plane z x y. Thus dv fx,yda x y dxdy x y dydx V x y da where is the base of the solid which is shown below. 5

6 The boundaries of are y,y x, andx. Hence x y V x y dydx xy y xdx x x x dx 3 x x dx 3 x 3 3 x. Note that another expression for the volume is V y x y dxdy. Properties of Double Integrals Double integrals have the same properties as integrals of one variable. For example, if c and c are constants, then 5 x x e 5 c fx,y c gx,yda c fx,yda c gx,yda. 4xe y 3ysin xdydx 3 3e 6sin 5e 5 7cos5e 5 6e 6 6e 7 6sine 5 9cose 5 whereas and x x 3 5 x x xe y dydx 8 e e 6 e 7 e 5 ysin xdydx 3sin5 7 cos5 3sin 9 cos 6

7 Adding the results given by and 3 gives after a bit of algebra. If is a closed region which can be decomposed into regions and and f is continuous over, then fx,yda fx,yda fx,yda. Let be the rectangular region x,y shown below consisting of the triangles and. We shall show that Now or x y 3 da x y 3 da x y 3 da x y 3 da x y 3 dxdy x y 3 da x y 3 dydx The triangle is given by x y, y so or x y 3 da y x y 3 dxdy 7

8 x y 3 da x x y 3 dydx Triangle is given by y x, x so or x y 3 da y x y 3 dxdy 8 x y 3 da x x y 3 dydx 8 Finally 8, which is the result we got before. Special Case-Area by Integration The special case of fx,ydx dx when f is da dxdy dydx. In this case the double integral represents the area of the region in the x,y plane. Area b f a x dy dx f x 8

9 d g Area c y dxdy g y The integral x x dy dx represents the area of a region of the x,y plane. Sketch the region and express the same area as a double integral with the order of integration reversed. Solution: The inner integral varies from y x to y x. Integral gives area of vertical strip between x and x dx for values of x fromto x Change order and take integration first. Then x goes from y to and y dy. Thus y to give a horizontal strip between y 9

10 x y x dydx dxdy. y y y dxdy y ydy y 3 3 y 3 6. Also x x dydx x x dx 6 which checks. Find the area bounded by the parabola y x and the line y x x Solution: The parabola and line intersect where y x x x x orx x Thus x, intersection. x y 4; whereasx y x are the x coordinates of the points of We shall first find the area as dxdy. When y is between and, x goes from y to y y dx dy y When y is between and 4, x goes from y to y 4 y dx dy. Thus y A y y dx dy 4 y dx dy. y We now set up the expression for the area the other way. dy dx x x dy dx y x x dx x x dx x x x3 3

11 Find the area between the parabola x y y and the line x y, that is the line y x x Solution: Now x y y or x y y. Completing the square ) x y y 4 4 or x 4 y. Hence the parabola passes through 4,.Nowx y y y and y. Thus the parabola goes through the points,,,, and 4,. We now find the points where the line and the parabola intersect. We have x y y and x y y y y or y y y y. y ory. The points of intersection are therefore, and,. We again set up the expression for area in two ways. First consider dxdy. y y A dxdy y y y ydxdy y y dy y y Now for y 4 x on the parabola. Thus dydx. A x 4 x dydx Change the order of integration in 4 cos x dx fx,ydy sinx 4 4 x 4 x dydx

12 dy sin y fx,ydx x dy cos y fx,ydx. Polar coordinates-change of variables ecall that given a point x,y we may assign to this point new coordinates r, as follows: x rcos y rsin tan y x r x y If r and are given, then they uniquely determine a point in the x,y plane. An equation of the form r f determines a curve in the x,y plane. This topic was discussed in Ma 6. Graph r sin 3 y x -. r sin3

13 Graph r 4cos y x - r 4rcos 4x. Sincer x y so that x y 4x or x 4x y, which is the circle x y 4 centered at, with radius. Area Using Polar Coordinates ecall da dx dy. Now from the figure ΔA r Δr Δ r Δ r Δrr Δr Δ r Δ ΔrrΔ Δr Δ rδrδ 3

14 fx,ydxdy da rdrd Fr,rdrd. Findtheareaofthecirclex y 4. Solution: We know that A r 4. Using double integration in polar coordinates, we have Evaluate e x y dx dy. A 4cos rdrd r 4cos d 6cos d 6 cos d 8 sin Switching to polar coordinates, we have e r cos r sin rdrd e r rdrd e r 8 4. d d 4 i Find the equations in polar coordinates of the curves x y y and x y x and graph the curves. Solution: The two curves are given by r sin and The graphs are given below. r cos 4

15 y - ii Give an integral or integrals in polar coordinates for the area between the two curves. Solution: A rdrd where is the region common to both circles. The two circles intersect when cos sin or when tan - x That is, at 4. r goes from to the circle r sin, for 4. Thus we need two integrals to express the area. 4 and from to r cos for A 4 sin rdrd 4 cos rdrd Find the area which lies inside the cardioid r a cos and outside the circle r a. Use double integration. The figure below shows the two curves with a. A rdrd a acos rdrd a 4 a 5

16 Give an integral in polar coordinates which represents the area of the region that lies outside the circle r a and inside the circle r a sin. Solution: We must sketch. First, x rcos, y rsin. Thus the circle r a is centered at the origin and has radius a. We rewrite the equation of the other circle. r arsin Thus x y ay or x y a a This circle passes through the origin, is centered on the y-axis at,a and has radius a. For convenience, a in the picture below To find the limits of integration, we have to equate the expressions for the two circles. a a sin sin 6, 5, where the circles intercept. So lies between these two 6 values. On the other hand, r goes from a to a sin for these values of. Theintegralfortheareais: a a sin rdrd 3 a a 3 Triple Integrals We shall now discuss a logical extension of the double integral. Consider V Fx,y,zdV V Fx,y,zdxdydz V This is clearly merely an extension of the double integral. Fx,y,zdydxdz Fx,y,zdzdxdy etc. V 6

17 Evaluate I x x xyzdzdydx x x x x I xyz dz dy dx xy dy xy x y x 4 y dx xx x dx 4 4 4x x 3x 3 6x 4 x 5 dx 3 4 dx 7

18 Compute the triple integral of Fx,y,z z over the region in the first octant bounded by the planes y, z, x y, y x 6 and the cylinder y z 4. 8

19 : y 4 y I zdv y zdzdxdy Volume The case f, i.e. dv is of particular interest. It yields the volume between two surfaces. To see this suppose a region of x,y,z space is bounded below by the surface z f x,y, above by the surface z f x,y and laterally by a cylinder C with elements parallel to the z axis. Let A denote the region of the x,y plane enclosed by the cylinder C. 9

20 Then the volume of the region is f V x,y dzdydx. f x,y A The x and y limits of integration extend over the region A. To get the x,y limits it is usually desirable to draw the x,y plane view of the solid. Often one can get the boundary of A by eliminating z from z f x,y and z f x,y, i.e.fromf x,y f x,y. Inthex,y plane this represents the boundary of A. Find the volume bounded by the paraboloid z x y and the parabolic cylinder z 4 y. x y Solution:

21 z: From paraboloid to cylinder x y 4 y y: Fromto x ; gotten by eliminating z from equations

22 x: Fromto set; y inx y V 4 x 4 y x y dzdydx 4 Find the volume of the solid region D between the parabolic cylinders z y and z y for x 3. Sketch D. Solution: y x 3.5 y ;y We obtain the intersection lines of the two surfaces: z z y y y. The limits of integration are then: x 3; y ; Then 3 y V dzdydx y y y dydx y dydx Cylindrical and Spherical Coordinates Cylindrical coordinates are related to Cartesian coordinates via x rcos y rsin z z The relationship between a volume element in the two systems is that is dv dxdydz rdrddz,

23 dv rdrddz Spherical coordinates are related to Cartesian coordinates via x,y,z,, where x cossin y sin sin z cos The relationship between a volume element in the two systems is dv dxdydz sinddd, that is dv sin ddd. It is important to keep in mind that is measured from the z axis and thus varies only from to. 3

24 Find the volume above the cone z x y and inside the sphere x y z az. 4

25 We shall use spherical coordinates. Cone: z x y z cos x cossin y sinsin The equation of the cone cos cos sin sin sin or cos sin tan 45 4 or Sphere: x y z az orx y z a a. Center at,,a. acos or a cos. We see that goes from to, fromto. and from to a cos. 4 Hence Volume sin dv 4 a cos sin ddd a 3 Give the expression in cylindrical coordinates for the volume of the solid inside both the cylinder x y 4 and the ellipsoid 4x 4y z 64. Sketch the volume. Do not evaluate this expression. SOLUTION 5

26 The ellipsoid intersects the x,y-plane in the circle x y 6. Thus, our region is bounded by the circle x y 4. So, in polar coordinates we have the equation r. Next, we can solve the equation of the ellipsoid 4x 4y z 64 for z, i.e., z x y 6 which can be rewritten in polar coordinates as z 6 r. The volume of the solid can now be written as: 6 r rdzdrd Additional Cylindrical and Spherical Coordinates s Give an expression in cylindrical coordinates for the volume of the solid T bounded above by the plane z y and below by the paraboloid z x y.sketcht. Donot evaluate this integral. y Solution: In cylindrical coordinates the plane has the equation z rsin and the paraboloid has the equation z r. The two surfaces intersect when y x y, that is the circle x y y or x y. This circle is only in first and second quadrants. The equation of this circle is 4 6

27 r sin in polar coordinates. y x sin r sin Volume r rdzdrd Evaluate cos x y z 3 dv V where V is the unit ball. Solution: V is given by x y z. In spherical coordinates the equation for V is. Thus V cos x y z 3 dv xyz V cos 3 sindv cos 3 sin ddd 4 3 sin.) Set up, but DO NOT INTEGATE, a triple integral to find the volume of the solid bounded above by x y z 5 and below by z using spherical coordinates. Solution: The region of integration is shown below. One uses Plot 3D, Implicit to get the picture. x y z 5 7

28 will go from the plane z to the sphere x y z 5. In spherical, x y z 5 5 Also, z cos sec. So, sec 5. For, we can form a right triangle with hypotenuse 5 (the radius of the sphere) and vertical side which is the distance from the origin to z. So the horizontal side is x z y Therefore, tan arctan. 8

29 So, arctan. The volume is: arctan 5 V p sin ddd sec Surface Integrals It is often necessary to integrate a function over a curved surface. Such integrals are called surface integrals. Let z x,y describe a particular surface S. Letf fx,y,z be a given function. We desire to integrate f over S, i.e.toevaluate S fx,y,zds fx,y,x,yds. S Here ds comes from dividing S into pieces ΔS. 9

30 The special case f is of particular interest, since ds area of curved surface S. S emark. The case, i.e. z corresponds to finding an ordinary double integral since then S is simply a region in the x,y plane. Question: How does one evaluate fds? S Suppose S is such that it can be uniquely projected onto the x,y plane. (We shall discuss the more general case later.) This is so if every line parallel to the z axis cuts S exactly once. As we take pieces ΔS smaller and smaller they approach flat pieces tilted with respect to the horizontal. Now if n is the normal to z x,y then n x i y j k. ecall if r is a vector from the origin to the surface then d r dx i dy j dzk. Then n d r x dx y dy dz. But z x,y dz x dx y dy n d r. Thus n is normal to the surface. Also n k n k cos where is the angle between the normal and the vertical. 3

31 cos n k n k x y x y. If da is the projection onto the horizontal plane ds of then da cos ds or ds cos da ds x y da x y dxdy x y dydx. Notice that da dx dy dy dx,, as expected. Hence S where is the projection of S onto the x,y plane. fx,y,zds fx,y,x,y x y da, Find the surface area of the paraboloid z x y below the plane z. Solution: The surface S projects into the interior of the circle x y. This is. Here 3

32 z x,y x y Surface area S ds x y da Here is circle x y. Thus the surface area is given by S 4x 4y dydx To evaluate this double integral we shall use polar coordinates. Then Check: Surface area 4r rdrd 8 4r 3 3 d 4r rdrd r rdrd 5 3 d 6 r 5 3

33 Evaluate I x y z ds S over the curved surface of the cone x y z which lies between z andz. Solution: Here z x,y x y. We need z x and z y.sincez x y zz x x z x x z and z y y z. Therefore Hence x y z x z y x z y z I S x y z ds x y x y x y z dxdy. is the interior of the circle x y. Again we use polar coordinates to evaluate I. Thensince z r on the cone we have I x y z ds r cos r sin r rdrd S 8 8 cos cos 4 drd cos4 4 d 8 cos cos 4 64 x cos4d 3. d Sketch the surface S given by the equation in the first octant. 3 x 3 y x y 3z 6 33

34 Find the surface area of S. Let S xy denote the projection of S onto the x,y plane. ThenS xy is the triangle shown in the first quadrant bounded by the line x y 6. 3 x y x Area Sxy z x z y da Sxy 3 3 da 4 3 Sxy da Find the surface area of the paraboloid given by z 4 x y for z. Sketch the surface. SOLUTION The surface is a paraboloid (only the part above the x,y plane): 34

35 - y The formula for its surface area is z x Since z 4 x y, z x z y z y da 4x 4y da da. Now the paraboloid intercepts the x,y plane, forming the circle x y 4. We have to determine the limits of integration of x and y using this circle. Introduce polar coordinates; then r goes from to and goes from to, and the surface-area integral becomes 4r rdrd