Solutions to Vector Calculus Practice Problems


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1 olutions to Vector alculus Practice Problems 1. Let be the region in determined by the inequalities x + y 4 and y x. Evaluate the following integral. sinx + y ) da Answer: The region looks like y y x x y x We use polar coordinates: π/4 π/4 π/4 π/4 r sinr ) dr dθ + 1 cos4) + 1 ) dθ + 5π/4 3π/4 5π/4 3π/4 r sinr ) dr dθ 1 cos4) + 1 ) dθ π cos4) + π 1
2 . Evaluate z dv, where is the following solid region: z,, 1) 1,, 1), 1, 1), 1, ),, ) y x 1, 1, ) Answer: If we integrate with respect to x first, we will not need to break the region into more than one piece that is, we will only need to use one integral). The plane containing the front parallelogram has equation x + y + z. You should be able to find this by the guessandcheck method.) o, the bounds for x are x y z. We are integrating over a square in the yzplane, so the y and z bounds are y 1 and z 1. Thus: 1 1 y z 1 1 z dv z dx dy dz ) z yz z dy dz y ) dz Let be the region in 3 determined by the inequalities r 1 and z 4 r, where r x + y. Evaluate r dv. Answer: is the region under a paraboloid, inside a cylinder, and above the xyplane. We will use cylindrical coordinates. π 1 4 r π 1 r dv r dz dr dθ r 4 r ) dr dθ π [ 4 3 r3 1 5 r4 ] 1 π 17 34π dθ 15 15
3 4. Let be the region in 3 defined by 4 x + y + z 9 and z. Evaluate the following integral: x + y ) dv Answer: is the region inside a sphere of radius 3, outside a sphere of radius, and above the xyplane. We will use spherical coordinates, because the region is very easy to describe in spherical coordinates: ρ 3, φ π, θ π. ince r ρ sin φ, we are integrating x + y r ρ sin φ. Thus: x + y ) dv π π/ 3 π π/ 3 π π/ 11 5 π π/ ρ sin φ ) ρ sin φ dρ dφ dθ ρ 4 sin 3 φ dρ dφ dθ 11 5 sin3 φ dρ dφ dθ We use the substitution u cos φ, du sin φ dφ: x + y ) dv 11 5 π 1 1 cos φ ) sin φ dφ dθ 1 u ) du dθ 844π Let be the curve x 1 y from, 1) to, 1). Evaluate y 3 dx + x dy Answer: We parameterize the curve using t y: x 1 t y t 1 t 1 3
4 Then: dx t dt dy dt Thus: y 3 dx + x dy 1 1 t 3 t) + 1 t ) ) 1 dt t 4 t + 1 ) dt 1 [ 1 5 t5 3 t3 + t ] Let be the curve x t, y 1 + t 3 for t 1. Evaluate x 3 y 4 dx + x 4 y 3 dy Answer: The rotation of this vector field equals : x y 4x 3 y 3 4x 3 y 3 x 3 y 4 x 4 y 3 This means that the vector field is conservative, so there is some function f with f x 3 y 4 i+x 4 y 3 j. It is fairly easy to see that f 1 4 x4 y 4 + works. The endpoints of the curve are, 1) when t ) and 1, ) when t 1). Thus: x 3 y 4 dx + x 4 y 3 dy [ ] 1,) 1 4 x4 y 4,1) 4 4 Note: This problem can also be done as a standard vector line integral, but the calculations are somewhat tedious. 4
5 7. Let be the circle x + y 4, oriented counterclockwise. Use Green s Theorem to evaluate the following integral cosx ) y 3) dx + x 3 dy Answer: First, we compute the rotation of the vector field: x y 3x + 3y cosx ) y 3 x 3 Then, by Green s Theorem: cosx ) y 3) dx + x 3 dy 3x + 3y ) da where is the region inside the circle. Using polar coordinates, we have: cosx ) y 3) π dx + x 3 dy 3r 3 dr dθ 4π 5
6 8. The following picture shows the parametric curve x, y) t t 3, t ): y, 1) x 1 1 Use Green s Theorem to find the area of the shaded region. Answer: First, we need to find the bounds for t. In particular, we want to know when the curve passes through the the point, 1). etting t t 3 and t 1 and solving, we get t 1 and t 1. Thus, our bounds are 1 t 1. By plotting a few points of the parametric curve, we can see that the curve is oriented counterclockwise, which agrees with the orientation given in Green s Theorem. We need to find a vector field F such that rotf) 1. The vector field F y i will work as would x j or lots of other possibilities). Then, by Green s Theorem: da y dx where is the shaded region and is the curve. We have: for 1 t 1. Thus: da x t t 3 dx 1 3t ) dt y t dy t dt 1 1 t 1 3t ) dt
7 9. Let be the surface given by the following parametric equations x 4t + u y cos t z sin u Find the equation for the plane tangent to the surface at the point π,, ). Answer: ince we are given the surface by parametric equations, we can find the normal vector by computing T t T u, and then we can use the normal vector to find the equation for the plane. T t 8t, sin t, ) T u 1,, cos u) i j k T t T u 8t sin t sin t cos u i 8t cos u j + sin t k 1 cos u We can determine the values of t and u at the point, 1, ). At the point π,, ), 4t + u π, cos t, and sin u. olving, we get t π/ and u. Thus, the normal vector at this point is: T t T u π, ) i 4π j + k Thus, the equation for the plane is of the form x 4πy + z D. We can solve for D by plugging in the point π,, ). We get that π D. Thus, the equation for the plane is x 4πy + z π 1. Let be the surface z x + y, z 4. Evaluate the following integral: x da z 7
8 Answer: The surface is a paraboloid. It will be easiest to work this problem if we use the parameters t r and u θ. Then, we can parameterize the surface as follows: x t cos u y t sin u z t t u π The tangent vectors to the surface are T t cos u, sin u, t) T u t sin u, t cos u, ) Then, the normal vector to the surface is i j k T t T u cos u sin u t t sin u t cos u t cos u i t sin u j + t k Thus: da T t T u dt du 4t 4 + t dt du t 4t + 1 dt du 4t 4 cos +4t 4 sin u + t dt du We can now compute the surface integral: x π t cos u da t 4t + 1 dt du z t π t cos u 4t + 1 dt du We use the substitution v 4t + 1, dv 8t dt: x da 1 π 17 cos u v dv du z 8 1 Note: It is also possible to see that the integral is without doing any computations, since the region is symmetric about the yzplane, as is the function x/ z. This is similar to how the integral of an odd function from a to a is. 8
9 11. Let be the surface z x + y, 1 z. Evaluate the following integral: x i + y j + z k ) da Answer: We will parameterize this surface with parameters t r and u θ. Then, the parametric equations are x t cos u y t sin u z t 1 t u π The tangent vectors to the surface are T t cos u, sin u, 1) T u t sin u, t cos u, ) The, the normal vector to the surface is i j k T t T u cos u sin u 1 t cos u i t sin u j + t k t sin u t cos u Now, we compute F T t T u ): F T t T u ) x, y, z ) t cos u, t sin u, t) t cos u, t sin u, t ) t cos u, t sin u, t) t cos u t sin u + t 3 t + t 3 Now, we can compute the surface integral: x i + y j + z k ) π da F T t T u ) dt du 1 π 1 t + t 3) dt du 17π 6 Note: This problem should have specified an orientation. The above answer corresponds to unit normals oriented pointing inwards towards 9
10 the zaxis you can see this from the vector T t T u ). The answer would be 17π if we oriented the surface with unit normals pointing 6 outwards. 1. Let be the union of the cylinder x + y 4 for 3 z 3 and the hemisphere x + y + z 3) 4 for z 3. Use tokes s Theorem to evaluate yz j z k ) da. Answer: In order to use tokes s Theorem, we need to compute the anticurl of yz j z k. The vector field F yz i will work. Then: yz j z k ) da yz i ds where is the boundary of the surface. The boundary of the surface is the circle x + y 4 in the plane z 3. We can parameterize this curve: x cos t y sin t t π z 3 Then: Thus: yz j z k ) da dx sin t dt dy cos t dt dz π 36π yz dx π 36 sin t dt 18 sin t sin t) dt π ) 1 cost) 36 dt Note: The region should have been oriented. The given answer is correct if the region is oriented with outward pointing normals. If the normals pointed inwards, the answer would be 36π. 1
11 13. Let be the the rectangle in 3 with vertices,, ), 1,, ), 1, 1, 1), and, 1, 1), oriented in the given order. Use tokes s Theorem to evaluate the following integral: sinx ) dx + xy dy + xz dz Answer: Using tokes s Theorem, we can change the line integral into a surface integral over a surface whose boundary is the given curve. One such surface is the interior of the given rectangle. The plane containing the four points has equation y z, so we can parameterize the surface using the parameters t x and u y. ince the rectangle lies over the unit square in the xyplane, the bounds for x and y are x 1 and y 1. x t y u z u t 1 u 1 Then, according to tokes s Theorem: sinx ) dx + xy dy + xz dz F) da where is the surface parameterized above and where F sinx ) i + xy j + xz k. We can compute F: i j k F z j + y k x y z sinx ) xy xz To compute the surface integral, we need to find T t T u : T t 1,, ) T u, 1, 1) o, T t T u, 1, 1). This vector points downwards, which disagrees with the orientation of the rectangle which is oriented counterclockwise 11
12 about upwards pointing vectors), so we need to negate the resulting integral. Thus: F) da z j + y k ), 1, 1) dt du z y ) dt du u ) dt du Let be the boundary of the region x + y 4, z 3, oriented with unit normals pointing outwards. onsider the vector field F x 3 + cosy ) ) i + yz j + 3y z + cosxy) ) k Use the divergence theorem to evaluate the following integral: F da Answer: The region is a cylinder. Note that the region is oriented appropriately to apply the divergence theorem. Let us denote the region by. Using the divergence theorem, we have F da F) dv We can compute the F: F 3x + z + 3y We can set up the integral in cylindrical coordinates: F da π 3 3r + z ) r dz dr dθ 9π 1
13 15. Let be the surface in 3 defined by the parametric equations r + cos t + cos u θ t z sin u t π u π Use the divergence theorem to find the volume of the region inside of. Answer: In order to use the divergence theorem, we need to compute the antidivergence of the constant function 1. ome simple vector fields that work are x i, y j, and z k. We will use z k, as it will make the computations easier. We need to parameterize the surface in terms of x, y, and z. We use the same parameters t and u as above, and the fact that x r cos θ and y r sin θ. Then: x + cos t + cos u) cos t y + cos t + cos u) sin t z sin u t π u π Note that the equation for z is simpler than the equation for x and y; this is why we are using the vector field z k. The tangent vectors to the surface are T t sin t cos u sin t, cos t + cos t + cos u cos t sin t, ) T u sin u cos t, sin u sin t, cos u) The normal vector to the surface is i j k T t T u sin t cos u sin t cos t + cos t + cos u cos t sin t sin u cos t sin u sin t cos u ince we will be computing T t T u ) z k, we only need to know the k component of T t T u. sin t sin u+sin t sin u cos u+ cos t sin u+cos 3 t sin u+cos u cos t sin u sin 3 u cos t 13
14 We can simplify this some using cos t + sin t 1. We get: Thus: sin u + cos 3 t sin u + cos u sin u sin 3 u cos t T t T u ) z k sin u sin u + cos 3 t sin u + cos u sin u sin 3 u cos t ) Thus: Volume π sin u + cos 3 u sin u + cos u sin u sin 4 u cos t π π π dv zk da π π z k T t T u ) dt du sin u + cos 3 u sin u + cos u sin u sin 4 u cos t ) dt du sin u + cos 3 u sin u + cos u sin u ) du To integrate the first term, we use the trig identity sin u 1 1 cosu)). To integrate the second term, we use the substitution v cos u, dv sin u du. To integrate the third term, we use the substitution w sin u, dw cos u du. π Volume π 1 cosu)) du + π 4π + + 4π 1 1 v 3 dv + π w dw ince the answer is positive, we know that the surface was oriented appropriately. 14
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