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Calculus III Practice Problems 9: Answers xy that lies inside the cylinder x 2 1. Find the surface area of the part of the hyperbolic paraboloid z Answer. Let R be the disk x2 y2 R. Now zx y zy x, so that Sur f ace Area 4. We want the surface area of the part of the surface z dS R R z 2 x 1 z2y dxdy R 1 y 2 x2 dA y 2 4. xy lying over It is convenient to switch to polar coordinates. We get 2π Sur f ace Area 2 0 0 1 2π 3 5 3 r2rdrd θ y2 2. Find the surface area of the part of the hyperbolic paraboloid z x2 y2 1 and x2 y2 4. 2π 0 2 1 1 4r2rdrd θ 2π x2 that lies between the cylinders Answer. We use the formula dS 1 z x 12 18 23 1 4r2 2 3 x 3. Find the surface area of the part of the surface z quadrant bounded by the line x y 1 . 2 x z2y dxdy: ∂z ∂x x 3 2 2 1 3 1 17 2 12 2y, dS y 5 1 4x2 1 1 x y that lies above the triangle in the first 3 2 3 2 y ∂z ∂y 1 0 1 3 2 1 x 0 0 3 2 x 1 0 y 1 x 0 z 1 x ydz dy dx Mass ρ dV The innermost integral is y 1 x y y 1 x y . Then y 1 x y dy y 1 x y 1 x 2 3 6 Answer. The solid can be represented by the inequalities, 0 0 1 x 0 1 x y 0 2 1 x 2 3 3 0 and the mass is Mass 1 6 1 x 1 0 3 dx 1 x 0 1 24 x, 0 4. Find the mass and the x-coordinate of the center of mass of the solid bounded by the planes x 0 z 0 x y z 1 with the density function ρ x y z y 1 4y2dA. 3 2 1 x ydxdy. The region is the type 1 domain given by the inequalities 0 x 1 0 y 1 1 x ydy dx 2 1 x y dx 2 2 dx 2 8 S 3 3 3 0 2 4 . Since z 2x z Answer. The surface lies over the region R 1 x 2 y2 Now, switching to polar coordinates, the surface area is so dS so 1 3 0 y y. Thus xydz dy dx The innermost integral is xy 1 x y x y 1 x y , and x y 1 x y dy x y 1 x y x 1 x 2 3 6 Now 1 Momx 0 1 x 0 1 x y 0 0 2 1 x 2 3 3 0 Finally, Momx using the substitution u 1 1 6 0 1 x dx 1 u u du 6 1 120 1 24 1 5. x. Thus x̄ 1 3 x1 0 3 1 x 0 3 1 120 4 0 5. Find the center of mass of the piece of the solid parabolic shell z 16 x 2 y lying above the xy-plane. 2 Answer. Since this is a solid of revolution about the z-axis, the center of mass lies on the z axis, so is of the form 0 0 z . To find z we must calculate the volume and Momz 0 for the region. For these calculations we switch to cylindrical coordinates. r r dr 2π 8r 4 Mom zdz rdrd θ π 16 r rdrdθ We conclude the computation with the change of variable u 16 r du 2rdr: 2 π Mom π u du 6 2π Volume 0 4 r rdrdθ 0 2π z 0 0 2π 2 16 4 3 16r 2 0 16 r2 4 0 4 4 4 0 27 π 2 2 0 0 2 16 z 0 Thus z 12 2 0 1 212 π 6 27 π 32 6 5 333 Answer. R can be described as the set of x y z 6. Find the average value of f x y z x y z over the region R in the first octant (the region where all the coordinates are positive) under the plane x y z 1. satisfying 0 description tells us how to calculate by iterated integrals: x 1 0 y 1 x 0 z 1 x y. This dz dy dx x y z dz dy dx Total f dV and the average is Total Volume. The computations are tedious and involve only integrations of polynomials. The final integrations (with respect to x) are Volume 1 2 x dx 16 Volume 1 1 x 1 x y dV R 0 1 R 0 1 x 0 0 0 1 x y 0 2 1 0 Total 1 0 1 3 x 2 x3 dx 6 1 8 Thus the average is (1/8)/(1/6) = 3/4. 7. The curve z x 1 2 0 z 1 is rotated about the z-axis, enclosing, together with the xy-plane, a 3-dimensional region R. R is filled with a substance whose density is inversely proportional to the distance from the z-axis. Find the total mass of this object. r 1 r 1 2π k 3 2 Answer. The region is that under the curve (using polar coordinates) z r 1 on the xy-plane. The density is δ k r. Thus, the mass is R zδ drd θ 2π 0 y 2 z2 1 2k r 0 8. Evaluate where R is the ball x2 r 1 R x 2 R π 2π 0 2π k 3 1 0 2 Answer. Switch to spherical coordinates. R is given by ρ ρ 2 dV 0 y z dxdydz 2 4. 3 rdr d θ above the disc R 0 2 0 2, and the integral is ρ 4 sin φ d ρ d θ d φ 4π 25 5 9. Find the centroid of the region R described in example 24, Chapter 17. 60. We used the change of variables Answer. There we found Volume u x x y z v y z u v y v w w z z w to do the computation. We had the Jacobian equal to 1, so it is easy to do the computation in the u v w variables: Mom x 0 0 3 5 xdV R 0 4 v dudvdw 4 u v dv du 5 3 u 0 0 0 The inner integral is v 3u 9 2 2 5 9 Mom 4 3u du 4 15 9 60 2 2 ydV 0 v w dudvdw 5 v w dv dw Mom 9 5 30 Mom 5 3w dw 9w 3w 2 2 zdV 0 wdudvdw 15 wdw 120 Mom 2 3 0 uv Thus 5 x 0 0 3 5 y 0 R 4 0 4 0 0 4 y 0 Finally 2 0 3 5 z 0 R 0 0 4 0 4 0 3 4 0 Thus the centroid is at (1,-1/2,2). For completeness, here is Example 24. Find the volume of the region R given by the inequalities 0 z 4 0 y z 3 0 x y z 5 This region is a parallelipiped, so by the appropriate change of coordinates, can be made to correspond to a rectangular parallelipiped. That is, we make the change of variables u x y z v y z w z u 5 0 v 3 0 w ∂ x y z dudvdw Volume dxdydz ∂ u v w Now, to calculate the Jacobian, we solve for x y z in terms of u v w: x u v y v w z w so that R corresponds to the region S given by the inequalities 0 R so that ∂ xyz ∂ uvw Thus S det 5 Volume 0 1 0 0 1 1 0 0 0 1 3 0 4 dudvdw 0 1 60 4. Thus