Math 211, Sections 02,04 Fall 2019 Solutions to Practice Problems 3

Math 211, Sections 02,04 Fall 2019 Solutions to Practice Problems 3 1. For the following integrals, sketch the region of integration, and then reverse the order of integration. Z 1 Z 1−x2 f (x, y) dy dx is bounded by y = 1−x2 above and by y = 1−x Answers. (a). The region for 0 1−x below. (The two curves intersect at (0, 1) and (1, 0); see me for a sketch.) Solving for x, those two √ curves can be written as x = 1 − y (to the left) and x = 1 − y (to the right). So the new integral is Z 1 Z √1−y f (x, y) dx dy. 0 1−y Z ln 3 Z 3 g(x, y) dx dy is bounded by x = ey to the left and above, by x = 3 to the (b). The region for 0 ey right, and by y = 0 below. The curve x = ey is of course y = ln x, which intersects y = 0 and (1, 0), Z 3 Z ln x and x = 3 at (3, ln 3). So the new integral is g(x, y) dy dx. 1 0 ZZ 2. Compute 2x + y dA, where D is the region in the plane bounded by the line y = 3 − 2x and D the hyperbola xy = 1. Answer. Rewriting the hyperbola as y = 1/x, we see that the two curves intersect when 1/x = 3−2x, i.e., 2x2 − 3x + 1 = 0, i.e., (2x − 1)(x − 1) = 0, i.e., x = 1/2 and x = 1. A sketch shows the line is on top, the hyperbola is below. So the integral is Z Z 1 Zand Z 1 3−2x 1 1 3−2x 1 1 2x + y dy dx = 2xy + y 2 dx = 6x − 4x2 + (3 − 2x)2 − 2 − 2 dx 2 y=1/x 2 2x 1/2 1/x 1/2 1/2   Z 1 1 5 1 2 3 5 1 1 5 1 2 5 1 2 = −2x + − 2 dx = − x + x + − −1 = =− + + + 2 2x 3 2 2x 1/2 3 2 2 12 4 6 1/2 ZZ 3. Compute x2 y dA, where D is the region in the first quadrant inside the circle x2 + y 2 = 4 and D outside the circle x2 + y 2 = x. Answer. In polar coordinates, the outer circle is r = 2, and the inner circle is r2 = r cos θ, or equivalently, r = cos θ. The two curves never intersect (since cos θ < 2 always); since we are restricted to the first quadrant, then, θ runs from 0 to π/2. Thus, the integral is Z π/2 Z 2 Z π/2 Z 2 2 (r cos θ) (r sin θ)r dr dθ = r4 cos2 θ sin θ dr dθ 0 θ 0Z cos θ Z cos π/2 5 2 1 π/2 r 2 = cos θ sin θ dθ = (32 cos2 θ − cos7 θ) sin θ dθ 5 5 r=cos θ 0 Z0 1 0 1 h 32 3 1 8 i0 1  32 1  [u = cos θ, du = − sin θ dθ] = − u − u = − . (32u2 − u7 ) du = − 5 1 5 3 8 5 3 8 1 [On an exam, you can leave a messy sum of fractions like that. (And I’ll make sure that it doesn’t get that bad anyhow.) But for the record, the above mess simplifies to 253/120.] ZZ 4. Compute 4 + x2 dA, where D is the region in the plane bounded by y = x2 + 1 and y = 3 − x2 . D Answer. The two curves intersect when x2 + 1 = 3 − x2 , i.e., 2x2 = 2, i.e., x2 = 1, i.e., x = ±1. A sketch shows y = 3 − x2 is on top, and y = x2 + 1 is below. So the integral is Z 1 Z 3−x2 Z 1 Z 1 3−x2 2 2 4 + x dy dx = (4 + x )y dx = (4 + x2 )(2 − 2x2 ) dx 2 +1 y=x 2 −1 Zx +1 −1 −1   1 2 51 2 2 56 2 4 3 = 8 − 6x − 2x dx = 8x − 2x − x =8−2− +8−2− = . 5 −1 5 5 5 −1 ZZ x3 + xy 2 dA, where D is the region in the first quadrant bounded by the circle √ x2 + y 2 = 1 and the lines y = x and y = 3x. 5. Compute D Answer. In polar√coordinates, the line y = x in the first quadrant is θ = π/4, and (since tan π/3 = √ 3), the line y = 3x in the first quadrant is θ = π/3. Meanwhile, x3 + xy 2 = x(x2 + y 2 ) = r3 cos θ. Thus, the integral is √ √ Z π/3 Z 1   √3 √2  1 π/3  r 5 1  3− 2 4 r cos θ dr dθ = sin θ = − ( − 0) = . 5 0 2 2 5 10 π/4 π/4 0 ZZ x2 + y dA, where D is the region in the plane bounded above by the curves y = x2 6. Compute D and y = 6 − x and below by the x-axis. Answer. [If we try to intergrate dx dy, the upper limit of integration on the inner integral would be a piecewise-defined function; so let’s integrate dy dx.] The two curves intersect when x2 = 6 − x, which is to say x2 + x − 6 = 0, i.e., (x − 2)(x + 3) = 0, so at x = 2 and x = −3. A sketch of the region, however, shows that the region in question is in the first quadrant, and hence the point of intersection has x-coordinate x = 2. However, we actually need the y-coordinate, which is y = 22 = 6 − 2 = 4. √ Since the curve y = x2 , which along the edge of this region is x = y, is the left boundary of the region, and y = 6 − x, which boundary, the integral is Z 4 Z 6−y Z 4 is3 x = 6 − y, is theZright 4 6−y x 1 x2 + y dx dy = + xy √ dy = [(6 − y)3 − y 3/2 ] + (6 − y)y − y 3/2 dy √ x= y 0 y 0 3 0 3 Z 4 4 1 8 1 4 1 (6 − y)3 − y 3/2 + 6y − y 2 dy = − (6 − y)4 − y 5/2 + 3y 2 − y 3 = 3 12 15 3 0 0 3   16 8 · 32 64 64 68 256 596 1744 2 =− − + 48 − + =− − + 48 + 3 · 6 = − + 156 = 12 15 3 12 3 15 15 15 7. Let D be the region in the plane that lies inside the circle x2 + y 2 = 6y but outside the circle x2 + y 2 = 9. Compute the area of D. Answer. The first circle is r2 = 6r sin θ, i.e., r = 6 sin θ; and the second is r = 3. They intersect when 3 = 6 sin θ, i.e., sin θ = 1/2, so θ = π/6 and θ = 5π/6. [And the region is definitely in the sector between these two angles, since the first circle is completely above the x-axis.] So the area is ZZ Z 5π/6 Z 6 sin θ Z 5π/6 2 Z 5π/6 r 6 sin θ 9 1 dA = r dr dθ = dθ = 18 sin2 θ − dθ 2 2 r=3 D π/6 3 π/6 π/6 √   √ √ Z 5π/6  5π/6 9 9θ 9 15π 9 3 3π 9 3  9 3 = = − 9 cos 2θ dθ = − sin 2θ + − − = 3π + 2 2 2 4 4 4 4 2 π/6 π/6 Z 1Z 1 cos(πy 4 /2) dy dx. 8. Compute 0 x1/3 Answer. The first integration looks hopeless, so we note that the region of integration is bounded on the left by x = 0, above by y = 1, and on the right (and below) by y = x1/3 , which is equivalently x = y 3 . The lowermost point of the region is the origin. Thus, the integral is Z 1 Z y3 Z 1 Z 1 y3 4 4 cos(πy /2) dx dy = x cos(πy /2) dy = y 3 cos(πy 4 /2) dy x=0 0 0 0 0 Z π/2 π/2 1 1 1 1 [u = πy 4 /2, du = 2πy 3 dy] = cos u du = sin u = (1 − 0) = . 2π 0 2π 2π 2π 0 9. Find the volume of solid that lies between the paraboloids z = x2 + y 2 and z = 36 − 3x2 − 3y 2 . Answer. In cylindrical coordinates, the first paraboloid is z = r2 , and the second is z = 36 − 3r2 . The surfaces intersect when r2 = 36 − 3r2 , i.e., r2 = 9, i.e., along the circle r = 3 [and z = 9]. Thus, if we squash the solid onto the xy-plane, we get the disk r ≤ 3. Above that disk, z = r2 lies below z = 36 − 3r2 . Hence, the volume is Z 3 Z 3 Z 2π Z 3 Z 36−3r2 3 36r − 4r3 dr = 2π(18r2 − r4 ) r(36 − 3r2 − r2 ) dr = 2π r dz dr dθ = 2π 0 r2 0 = 2π · 9 · (18 − 9) = 162π. 0 0 0 10. Find the volume of the solid bounded by the surface z = ex and the planes x = y, x = 1, y = 0, and z = 0. Answer. Z Squashing onto the xy-plane the triangle Z 1 bounded by x = y, x = 1, and y = 0. So the Z 1 Z gives x 1 Z x Z ex xex dx [u = x, dv = dx] ex dy dx = dz dy dx = volume is 0 0 0 0Z 0 0 h i1 1 1 = xex − ex dx = (e − 0) − ex = e − e + 1 = 1. 0 0 0 11. Let E be the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 2, 0), and (0, 0, 4), and with constant density 1. Find the mass and center of mass of E. Answer. The four faces of the tetrahedron are the coordinate planes and the plane through the last three points. Taking the vectors between pairs of these points, we see that h−1, 0, 4i and h0, −2, 4i lie in this plane. Taking the cross product shows that h8, 4, 2i, and hence h4, 2, 1i, is normal to this plane. So the plane has equation 4x + 2y + (z − 4) = 0, i.e., z = 4 − 4x − 2y. Squashing the tetrahedron onto the xy-plane Z Z gives Z the triangle bounded Z by Z x = 0, y = 0, and y = 2 − Z 2x. Thus, the mass is 1 2−2x 4−4x−2y 1 2−2x 1 2−2x (4 − 4x)y − y 2 dx y=0 0 0 0 Z 0 Z 01 0  1 1 4 4 4 (4 − 4x)(2 − 2x) − (2 − 2x)2 dx = 4 (1 − x)2 dx = − (1 − x)3 = − 0 − = = . 3 3 3 0 0 0 Next, the yz-moment is Z 1 Z 2−2x Z 4−4x−2y Z 1 Z 2−2x x dz dy dx = 4x − 4x2 − 2xy dy dx 0 0 0 0 Z0 1 Z 1 1 2−2x 8 2 2 = 4xy − 4x y − xy dx = · · · = 4x3 − 8x2 + 4x dx = x4 − x3 + 2x2 3 y=0 0 0 0 8 1 =1− +2= . 3 3 TheZxz-moment is Z 1 Z 2−2x Z 1 Z 1 2−2x Z 4−4x−2y 2 2−2x y dz dy dx = 4y − 4xy − 2y 2 dy dx = (2 − 2x)y 2 − y 3 dx 3 y=0 0 0 0Z 0 0 0 1 8 1 2 2 2 = ··· = (1 − x)3 dx = − (1 − x)4 = 0 + = . 3 0 3 3 3 0 Last, the xy-moment is Z 1 Z 2−2x Z 4−4x−2y Z Z 1 1 2−2x z dz dy dx = (4 − 4x − 2y)2 dy dx 2 0 0 0 0 0 Z 1 Z 1 y 3 2−2x 16 1 4 4 2 2 ··· = 2 4(1 − x) y − 2(1 − x)y + dx = (1 − x)3 dx = − (1 − x) = . 3 y=0 3 0 3 3 0 0 Therefore, the mass is 4/3, and the center of mass is 3/4(1/3, 2/3, 4/3) = (1/4, 1/2, 1). m= 4 − 4x − 2y dy dx = dz dy dx = 12. Let E be the solid region below the plane Z Z Z z = 10 − x, above the plane z = x + y, and between the surfaces x = y 2 and y = x − 2. Compute y dV . E Answer. The last two surfaces are vertical and give the region in the xy-plane with the (sideways) parabola x = y 2 to the left, and the line x = y + 2 to the right. (The intersection points are (4, 2) and (1, −1).) Meanwhile, the two original planes intersect above the line y = 10 − 2x, which, as can be seen from a sketch, is well outside the region just described. (That is, the two planes don’t further cut down the region in the xy-plane.) So if we squash the solid to the xy-plane, we get the 2D region between x = yZ2 and So the integral is Z Z x =Zy + 2 that we just Zdescribed. ZZZ 2 y+2 10−x 2 y+2 (10 − 2x − y)y dx dy y dz dx dy = −1 y 2 −1 y 2 x+y Z 2 Z E2 y+2 (10y + 20 − y 2 − 4y − 4 − y 2 − 2y − 10y 2 + y 4 + y 3 )y dy dy = (10x − x2 − xy)y = x=y 2 −1 −1 Z 2 2 y6 y5 4y 3 y 5 + y 4 − 12y 3 + 4y 2 + 16y dy = = + − 3y 4 + + 8y 2 6 5 3 −1 −1   32 32 32 1 1 4 68 33 1 81 = + − 48 + + 32 − + + 3 + − 8 = + − − 21 = 3 5 3 6 5 3 3 5 6 10 y dV = 13. Let E be the solid region between the parabolic cylinder y = 2z 2 and the y = −2z, bounded Z Zplane Z 2 in front by the plane x = 4 and in back by the plane x + z = y. Compute z dV . E Answer. Given that the first two surfaces described are both cylinders running parallel to the xaxis, it makes sense to squash the solid onto the yz-plane. In the yz-plane, the curves y = 2z 2 [which is a parabola opening along the positive y-axis] and y = −2z enclose a region in the third quadrant, extending diagonally from the origin down to the point (y, z) = (2, −1) where the two curves intersect. In front of this region, the plane x = 4 always lies in front of the plane x + z = y [because, in particular, x = y − z attains its maximum value on the region at the corner (2, −1), where x = y − z = 2 − (−1) = 3 ≤ 4]. Thus, the integral is Z 0 Z −2z Z 4 Z 0 Z −2z Z 0 Z −2z 4 2 2 z dx dy dz = xz dy dz = 4z 2 − yz 2 + z 3 dy dz x=y−z 2 2 2 −1 2z −1 2z −1 2z Z 0 y−z Z 0 −2z 1 2 2 2 3 4 3 = 4yz − y z + yz −8z − 8z − 2z 4 + 2z 6 − 2z 4 − 2z 5 dz dz = 2 2 y=2z −1 Z−1 0 12 1 2 0 12 1 2 3 4 5 6 = −8z − 12z − 2z + 2z dz = −2z 4 − z 5 − z 6 + z 7 =2− + + . 5 3 7 5 3 7 −1 −1 As before, on an exam, leave a messy sum of fractions like that; but for the record, this one is 23/105. 2 2 2 14. Let EZbe Z Z the solid lying inside the sphere x + y + z = 4 and below the cone z = Compute z 2 dV . p x2 + y 2 . E Answer. In spherical coordinates, the sphere is ρ = 2, and the cone is φ = π/4. The solid E is the full portion the cone removed. Thus,Z Z Z Zball with the Z 2π Z π Z above  2π  Z π  Z 2  2 2 2 2 2 z dV = (ρ cos φ) ρ sin φ dρ dφ dθ = dθ cos φ sin φ dφ ρ4 dρ E 0 π/4 0 0 π/4 0 and substituting u = cos φ, du = − sin φ dφ in the second integral, we get  Z −1  ρ5 2   u3 √2/2  32  2√2  16π √ 64π = (2π − 0) − √ u2 du = 2π · = − (−1) = ( 2 + 4). 5 0 3 −1 5 15 8 15 2/2 15. Let E be the portion of the solid in the first octantZ Z bounded by the xz-plane, the yz-plane, the Z 2 2 plane z = 4, and the paraboloid z = x + y . Compute x dV . E Answer. z = x2 + y 2 is a circular paraboloid (i.e., bowl) opening upward from the origin. So with the plane z = 4, we get a filled-in bowl of height 4. (The paraboloid is the bottom, and the plane is the top.) Since E is in the first octant, E is just the front right quarter of this solid bowl, with the back and left faces being the yz- and xz-planes. Let’s use cylindrical coordinates, so that the paraboloid becomes z = r2 , and the plane remains z = 4. The two surfaces intersect when r2 = 4, i.e., r = 2. So the projection of E onto the xy-plane is the quarter disk (in the first quadrant) of radius 2 centered at the origin. The integral is therefore Z π/2 Z 2 Z 4 Z π/2 Z 2 Z 4 ZZZ r2 cos θ dz dr dθ (r cos θ)r dz dr dθ = x dV = 2 2 0 0 r 0 r E Z π/2  Z π/2 Z 2 Z π/2 Z 2 0 4 2 4 3 1 5 2 4 2 (4r − r ) cos θ dr dθ = r z cos θ dr dθ = = dθ r − r cos θ 3 5 z=r2 r=0 0 0 0         Z0 π/2 0 π/2 32 32  1 1 1 1 64 1 1 = = 32 sin θ (1 − 0) = 32 = . − cos θ dθ = 32 − − − 3 5 3 5 3 5 3 5 15 0 0 2 2 2 2 2 2 16. Let E be the solid region between the pspheres x + y + z = 1 and x + y + z = 2, and inside 2 2 the (upper half of the double) cone z = x + y . The density of E at (x, y, z) is z 2 . Compute the mass of E. √ Answer. In spherical coordinates, the spheres cut out 1 ≤ ρ √ ≤ 2, and the cone forces 0 ≤ φ ≤ π/4. ZZZ Z 2π Z π/4 Z 2 Recalling z = ρ cos φ, the mass is z 2 dV = (ρ cos φ)2 ρ2 sin φ dρ dφ dθ E 0 0 1 √   Z π/4  Z 2 ρ4 dρ [u = cos φ, du = − sin φ dφ] = 2π cos2 φ sin φ dφ 1 0 √ √ √ Z    ρ5 2  2/2 2π  u3 2/2  √ 2 = −2π =− 4 2−1 u du 5 3 1 1 √ 5 1  √ √ π √ π 2π √ 2 (4 2 − 1) 1 − = (4 2 − 1)(4 − 2) = (17 2 − 12). = 15 4 30 30 17. Let a > 0. Find the volume of the solid inside both the sphere x2 + y 2 + z 2 = a2 and the cylinder x2 + y 2 = ax, and in the first octant. Answer. [Note: it’s perhaps tempting to use spherical coordinates here, but the cylinder will make that a mess, so I’ll use cylindrical.] Converting to cylindrical coordinates, the equation for the cylinder is r2 = ar cos θ, i.e. r = a cos θ; note that is swept out for√only 0 ≤ θ ≤ π/2. [It’s not 0 to π or 0 to 2π.] Meanwhile, the sphere is r2√+ z 2 = a2 , i.e., z = ± a2 − r2 ; but because we’re in the first octant, actually z goes from 0 to the volume is + a2 − r2 . So √ Z π/2 Z a cos θ Z a2 −r2 Z π/2 Z a cos θ p r dz dr dθ = r a2 − r2 dr dθ [u = a2 − r2 , du = −2r dr] 0 0 0 0 0 Z Z 2 2 Z Z 1 π/2 a sin θ √ 1 π/2 3/2 a2 sin2 θ a3 π/2 =− dθ = u du dθ = − u (1 − sin3 θ) dθ 2 2 0 3 3 u=a 2 a 0 0 Z iπ/2 a3  π 2  a3 π/2 a3 h 1 2 3 = (1 + cos θ sin θ − sin θ) dθ = θ − cos θ + cos θ = − . 3 0 3 3 3 2 3 0 18. Let E be the solid bounded by the surfaces y = 1 − x2 , y = 0, z = y, and z = 1. Suppose that the density of the solid is ρ(x, y, z) = 2(1 + x4 ). Compute the mass of E. Answer. The projection onto the xy-plane is the region D lying between the parabola y = 1 − x2 and the line y = 0; so that’s a hump resting on the x-axis, from −1 to 1, with its peak at (0, 1). Note that y ≤ 1 at every point of D, so that the surface z = y lies below z = 1 over each point (x, y) of D. Thus, the mass is ZZZ Z 1 Z 1−x2 Z 1 Z 1 Z 1−x2 1 4 ρ dV = 2(1 + x ) dz dy dx = 2(1 + x4 ) dy dx E −1 0 y −1 0 z=y 1−x2 1
(1+x4 ) 2−2x2 −(1−x2 )2 dx dx = (1+x )(2y−y ) y=0 −1 Z 0 Z −1 Z−1 1 1 1 1 − x8 dx (1 + x4 )(1 − x4 ) dx = (1 + x4 )(1 − 2x2 + 2x2 − x4 ) dx = = −1 −1 −1  x9 1 1  1 8 8 16 =x− = 1− − −1+ = + = . 9 −1 9 9 9 9 9 Z 3 Z √9−y2 Z √9−x2 −y2 p 19. Compute y 2 x2 + y 2 + z 2 dz dx dy. √ Z = 1 Z Z 4 2(1+x )(1−y) dy dx = −3 − 0 1 4 2 1−x2 Z 9−x2 −y 2 Answer. The region of integration is the front half of the sphere of radius 3 centered at the origin; that is, the p hemisphere with x ≥ 0. Converting to spherical coordinates, recalling that y = ρ sin φ sin θ and ρ = x2 + y 2 + z 2 , the integral is Z π/2 Z π Z 3  1 3   1 Z π/2  Z π 5 3 2 sin φ(1 − cos2 φ) dθ ρ sin φ sin θ dρ dφ dθ = 1 − cos 2θ dθ ρ6 2 −π/2 6 0 0 −π/2 0 0 θ 1  π π  2 2  243  π/2  π  243  1 = − sin 2θ = + + = 81π. − cos φ + cos3 φ 2 4 3 2 4 4 3 3 2 −π/2 0 2 2 2 20. Let E be the solid between the spheres x2 + y 2 + z 2 = 1 and Z ZxZ + y + z = 9, below the cone p p z = x2 + y 2 , and above the cone z = − 3x2 + 3y 2 . Compute x2 dV . E Answer. √We use spherical coordinates. The first cone is z = r, which means φ = π/4; the second is z = − 3r,√which means φ = 5π/6, since that is the unique angle between 0 and π for which tan(φ) = −1/ 3. [Note that tan φ = r/z.] Meanwhile, θ goes all the way around, from 0 to 2π, and ρ goes from the inner radius of 1 to the outer radius of 3. Thus, the integral is Z 2π Z 5π/6 Z 3 Z 2π Z 5π/6 Z 3 2 2 2 3 (ρ sin φ cos θ) ρ sin φ dρ dφ dθ = cos θ dθ sin φ dφ ρ4 dρ 0 π/4 1 0 π/4 1 Z 2π Z 5π/6 h ρ 5 i3 1 2 = (1 + cos 2θ) dθ (1 − cos φ) sin φ dφ 2 5 1 0 π/4 [On the second integral, substitute u = cos φ, du = − sin φ] Z −√32 h i h u3 i−√3/2 2π 1 θ 1 242 2 + sin 2θ [2π − 0 + 0 − 0] · −u √ = · √ (u − 1) du · [243 − 1] = 5 2 4 10 3 2/2 0 2/2 √ √ √ √ i h √ √ √ √ 242π 3 3 2 2 242π 121π = − + − + = (10 2 + 9 3) = (10 2 + 9 3). 5 8 2 12 2 120 60 p 2 + y 2 and the cone z = 2 x2 + y 2 in the first 21. Let E be the Zsolid between the paraboloid z = x ZZ octant. Compute y 2 z dV . E Answer. Converting to cylindrical coordinates, the two surfaces are z = 2r and z = r2 . The paraboloid lies under the cone, until they intersect along the circle r = 2, z = 4. [Well, really for r2 = 2r, so at both r = 0, which is at the origin, and at r = 2, which gives z = 4.] Meanwhile, being in the first octant, θ runs from 0 to π/2. So in cylindrical coordinates, the integral is Z π/2 Z 2 Z 2r Z π/2 Z 2 Z 2r 2 (r sin θ) zr dz dr dθ = r3 z sin2 θ dz dr dθ 2 2 0 0 r 0 0 r Z π/2 Z 2 Z π/2 Z 2 1 3 2 2 2r 1 = r z sin θ dr dθ = (2r5 − r7 ) sin2 θ dr dθ 2 2 2 z=r 0 0 Z0 π/2 0 h1 i2 h θ 1 iπ/2  64  1 1 = (1 − cos 2θ) dθ r6 − r8 = − sin 2θ · − 16 2 3 16 0 2 4 3 0  16 0π 4 = −0−0+0 · = π. 4 3 3