Calculus III Practice Problems 8: Answers 1. What is the mass of the lamina bounded by the curves y is δ x y xy? 3x and y 6x x 2 where the density function Answer. Let R represent the region covered by the lamina. x PSfrag replacements 3 From the figure we see that R is a type 1 domain, bounded above by y 6x the range 0 x 3. (To see that, solve 6x x 2 3x for x). Thus the mass is Mass R 0 x 2 and below by y 3x over 6x x2 3 δ dA xydy dx 3x The inner integral is 6x x2 xydy x 3x y2 2 x 12x which is 81 2 21 20 42 525. Thus Mass 1 2 3 5 x 36x2 2 6x x2 3x 4 12x3 1 x6 2 6 27x3 dx 0 x 4 9x2 12x5 5 27x4 4 1 5 x 2 12x4 81 9 2 6 3 0 27x3 36 5 27 4 2. A lamina filled with a homogeneous material (the density is identicaly equal to 1) is in the shape of the region 0 x π 0 y sin x. Find its center of mass. Solution. π Mass 0 sin x dydx 0 π sin xdx 0 1 ydydx 2 π 8 and the center of mass is π 4 π 8 . π Thys ȳ 0 0 π sin x 0 π 2. To find ȳ, we calculate Now, by symmetry, the x-coordinate of the center of mass is x̄ Momy 2 0 sin2 xdx π 4 3. The surface H, given in cylindrical coordinates by z 2θ is a helicoid. What is the volume of the region R bounded above by H 0 θ 2π , below by the plane z 0 and lying over the disc r 1? Answer. Here we use polar coordinates. The volume is zdA R R 2θ rdrd θ 2π 1 2 0 0 rθ rdrd θ 2 1 rdr 0 2π θ dθ 0 2π 2 4. A beach B is shaped in the form of a crescent. We model this on the area between the circle of radius 1, centered at the origin, and the circle of radius 3/4 centered at the point (3/4,0), where the units are in miles. Suppose that the human density σ decreases as we move from the beach according to σ x y 1000 x 2 y2 2 people per square mile.What is the population on that beach? Answer. We move to polar coordinates r θ Then the crescent is the domain bounded by the circles r 1 and r 3 2 cos θ . These curves intersect when cos θ 2 3, let α represent those angles. Then B is given by the relations α θ α 1 r 3 2 cos θ , and the population is the integral of the population density on this beach: 3 cos θ α 2 103 Population σ dA rdrd θ r4 α 1 The inner integral is 3 2 cos θ 1 Then Since cos α 2 3 tan α 2 1 r 2 1 4 sec θ dθ 9 103 dr r3 103 2 Population 2 103 α 2 α 3 2 103 1 2 cos θ 1 103 θ 2 4 sec2 θ 9 4 tan θ 9 α 5, and the population is 397.5 people. 4 103 tan α 9 α 5. 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 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 6. As u v runs through the region u 2 r 1 1 0 v2 2k r r 1 2π k rdr d θ 3 3 1 0 2 above the disc R 0 2π k 3 1, the vector function u v I u v J 2 Xuv 2 2 2 uvK describes a surface S in three space. Write down the double integral which must be calculated to find the surface area of S. Answer. We have to integrate dS Xu X 2uI u Xv d udv: 2uJ vK Xv 2vI 2vJ uK 2v 2u J 4uvK and the ensuing calculation leads to X X ! 2 2 u v Thus S 2 2 u v dudv where R is the unit disc. We now switch to polar coordinates: u v r cos θ sin θ r cos2θ , so cos 2θ r drdθ 2 cos 2θ dθ 4 2 " cos 2θ dθ S 2 2 2 The last equality comes from the observation that the integral of cos 2θ around the full circle is 8 times the integral of cos 2θ through π 4 radians. Thus 1 2 S 4 2 # 2 2 2 Xu 2v 2 Xv 2u2 I u 2 2 2 v 2 2 2 R 2 2π 1 0 2 2 2π 3 0 2 2 2 π 4 0 0 $ 7. Find the volume of the region lying above the disc x 2 z sin π x2 y2 2 . y2 1 in the xy-plane, and below the surface sin π r 2 lying Answer. Switching to polar coordinates, this is the volume bounded above by the surface z above the disc r 1. Thus V 2π zdA R 0 This integral we calculate by parts; 1 0 1 0 sin π r 2 rdrd θ 2π 1 0 r sin π r 2 dr 2 cos π r 2 dr π 2 2 2 2 1 sin π r 2 π π π 1 π % r sin π r 2 dr 2 r cos π r 2 π 1 1 0 0 1 0 8. Find the mass of the lamina of the region R lying between the ellipses x 2 the density function is δ x y x 2 y2 . 4y2 1 and x2 4y2 4, where Answer. Make the change of coordinates u x v 2y. Then R corresponds to the region S in uv-space bounded by the circles of radius 1 and 2. Writing x and y in terms of u and v by x u y v 2, we calculate the Jacobian: ∂ xy 1 1 0 det 0 12 ∂ uv 2 & ' Thus ∂ x y Mass δ dA x y dxdy x y dudv ∂ u v 1 v dudv u 2 4 Now, we switch to polar coordinates in u v space, obtaining 1 r cos θ 1 r sin θ rdrdθ Mass 2 R 2 2 R S 2 2 S 2π 2 0 2 1 2 2 2 4 2 2 Integrating first with respect to θ , since 2π 0 cos2 θ d θ we finally obtain 2π 0 sin2 θ d θ π 5π 2 3 5π 16 1 75π r dr 8 1 8 4 32 We note that we could have made just one coordinate change, directly from x y to r θ : Mass x r cos θ r sin θ 2 y but doing it in two steps is conceptually clearer and computationally easier. 9. Find the area of the region R in the first quadrant bounded by the curves y 2 2x y2 5x x2 4y x2 Answer. Make the change of coordinates y2 x u x2 y v u 5 4 v 10. Then ( dxdy ( ∂ x y dudv Area R ∂ u v To calculate the Jacobian, we have to solve for x y in terms of u v. After a little bit of algebra, we find x u " v " y u " v " so that R corresponds to the region S in uv-space given by the inequalities 2 R 1 3 2 3 S 2 3 1 3 Now, computing the partial derivatives, we find " " u " v " " " u " v " ' ∂ x y 1 Area R ∂ u v dudv 3 ∂ xy ∂ uv Finally det & 1 2 3 2 3 v 3u 2 1 3 1 3 u v 3 2 1 3 1 3 3 1 2 3 2 3 3 1 9 5 S 2 4 9 13 10 dudv 4 6 10y.