Calculus IV Math 2443 Review for Exam 2 on Mon Oct 24, 2016 Exam 2 will cover This is only a sample. Try all the homework problems.

Transcription

1 Calculus IV Math 443 eview for xam on Mon Oct 4, 6 xam will cover This is only a sample. Try all the homework problems. () o not evaluated the integral. Write as iterated integrals: (x + y )dv, where is the domain made of the cylinder bounded by y x and x + y, cut by two planes z and z x + y +. () o not evaluate the integral. Change the order of integration: y y f(x, y) dx dy (3) Find the volume of the solid under the cone z x + y and above the ring 4 x + y 5. Use double integral and polar coordinate. (4) Let be the region bounded by the paraboloid y x + z and the plane y 4. Write as an iterated integral for x + z dv. (5) Set up an integral expression for the mass of the solid, whose density is ρ(x, y, z) x, in the first octant bounded by the cylinder x + y and the planes y z, x, and z. (6) escribe the region whose area is given by +sin θ r dr dθ. (7) Sketch the solid whose volume is given by the iterated integral: (8) valuate the integral using spherical coordinate: x x y (x + y + z ) dz dy dx. y 4 y dxdzdy. (9) Find the volume of the region bounded by the paraboloids z x + y and z 36 3x 3y. () Use change of variable to evaluate (x + y) da, where is the region bounded by the lines x + y, x + y, x y, x y 3. () Use change of variable: [x u, y 3v] to evaluate x da, where is the region bounded by the ellipse 9x + 4y 36.

2 () Let f(x, y) be a function defined on a rectangular region [a, b] [c, d]. efine the double integral f(x, y) da. (3) Let be a domain in the xy-plane given by a x b y g (x), and f(x, y) be a continuous function on. How is defined)? f(x, y) defined (after () is (4) Use Fubini s Theorem to prove: Let be a domain in the xy-plane given by a x b y g (x), and f(x, y) be a continuous function on. Then b g (x) f(x, y) f(x, y) dy dx. a (5) valuate x dv, where is the solid enclosed by the ellipsoid a Use the transformation x aρ sin φ cos θ y bρ sin φ sin θ z cρ cos φ + y b + z c. Use the Jacobian for the spherical coordinate and the properties of the determinant: If a column or row contains a common factor, then the factor comes outside the determinant.

3 3 4 3 () omain is - - The solid is a cylinder with well-defined lower surface z, and upper surface z x + y +. The base domain in xy-plane is bounded by the two curves y x and x+y. To get the end-points for x, you need to solve the simultaneous equation y x and x + y to get (, ), (, 4). The solid is Therefore, (x + y )dv z x + y + x y x x. x+y+ x x+y+ x (x + y )dz da (x + y )dz dy dx. () y x 4 x f(x, y) dx dy f(x, y)dy dx + f(x, y)dy dx. y x x (3) The height function is f(x, y) x + y. The volume is x + y da, where is the ring 4 x + y 5. With polar coordinate, this ring is simply r 5.

4 4 Thus the volume is x + y da π. r r dr dθ r dr dθ (4) x + z dv 4 x + z dy da x +z x + z (4 x z )da where is the disk on xz-plane surrounded by the circle x + z 4. Using polar coordinate x r cos t z r sin t we have Integral r(4 r ) r dr dt 8π. (5) The solid is a cylinder parallel to the z-axis. The lower surface is z, upper surface is z y. Mass y xdz da xyda,

5 where is a quarter of the disk in xy-plane surrounded by x + y. Using polar coordinate, Mass 4 8 (r sin t)(r cos t)r drdt sin t cos t dt sin t dt 6 cos t ] π (6).5 To understand the region in polar coordinate, draw the inner curve r (half of the circle of radius, centered at (, ), from (, ) to (, )), and r +sin θ (this is a curve starting at (, ), to (, ), then finally to (, )). The region is the part in between the two curves, and above the x-axis because θ π. So, if we call the region in between the circle and the second curve, the integral is exactly da, the area of that region. The r factor in the integral is the Jacobian for the polar coordinate (7) The solid is a cylinder over the triangle in yz-plane with lower surface x, and upper surface x y. (8) The body is exactly the part in the first octant of the solid ball of radius, centered at the origin. The integrand is (x +y +z ) ρ 4, and Jacobian is ρ sin φ. Therefore, the integral -.5

6 6 is ρ 4 ρ sin φ dρ dφ dθ π 4. (9) The upper surface is z 36 3x 3y and lower surface is z x + y. So volume is 36 3x 3y x +y dz da (36 4(x + y )) da, where is the projection of the region bounded by the intersection curve of the above two surfaces. That means we solve them simultaneously to get x + y 9 (and z 9). Thus is the disk x + y 9. Using polar coordinate, this integral is 3 (36 4r ))r dr dθ. () We use the change of variable This becomes u x + y v x y x (u + v) 3 y (u v) 3 Then Jacobian is The region becomes So the integral becomes 3 (x, y) (u, v) x + y x y 3 u u 3. u du dv dv 3.

7 () The Jacobian is 6. So, Integral 4u 6 da where is the region in uv-plane surrounded by the curve 36u + 36v 36 (which is the circle u + v ). Now, use polar coordinate: Integral 4u 6 da 4(r cos θ) r dr dθ 4r 3 cos θ dr dθ cos θ dθ ( + cos θ)dθ (θ + sin θ) ] π π. 7 () ivide the intervals [a, b] and [c, d] into m, n equal subintervals. Let x b a m A x y. Pick a sample point (x i, yj ) from (i, j)th rectangle ij. Then m n f(x, y) da f(x i, yj ) A if the limit exists. lim m,n i j, y d c n ; (3) Find c < d such that c g (x) d for all a x b. So, [a, b] [c, d]. We also extend the function f(x, y, z) to F (x, y, z) by { f(x, y), if (x, y) F (x, y), if (x, y) Then f(x, y) da is defined by F (x, y) da. (4) Find c < d such that c g (x) d for all a x b. So, [a, b] [c, d]. Then f(x, y) da is defined by F (x, y) da, where { f(x, y), if (x, y) F (x, y), if (x, y)

8 8 Therefore, (F) Now, d c F (x, y) dy f(x, y) da g (x) c g (x) c g (x) b d a F (x, y) da c F (x, y) dy + dy + g (x) f(x, y) dy. Thus, from equality (F), we have f(x, y) da F (x, y) dy dx (by Fubini) g (x) b g (x) a F (x, y) dy + f(x, y) dy + d g (x) f(x, y) dy dx. (5) (Properties of eterminant) For example, cx cx cx 3 x x x 3 x 3 x 3 x 33 c x x x 3 x x x 3 x 3 x 3 x 33. d g (x) dy F (x, y) dy Since the Jacobian of the spherical coordinate is ρ sin φ, Jacobian for our coordinate is J abcρ sin φ, and the equation of the ellipse in this new coordinate is simply ρ. Thus, dv abcρ sin φ dρ dφ dθ 3 abc [ 3 abc cos φ 3 abc 4π 3 abc. sin φ dφ dθ Note that, if a b c r (solid ball of radius r), the the volume would have been 4 3 πr3. dθ ] π dθ