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
. 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 guessand-check method.) o, the bounds for x are x y z. We are integrating over a square in the yz-plane, 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 1 3 1 y ) dz 5 1 3. 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 xy-plane. 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
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 xy-plane. 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 15 5. 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
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 ] 1 1 4 15 6. 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
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
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 8 15 6
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
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 yz-plane, as is the function x/ z. This is similar to how the integral of an odd function from a to a is. 8
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
the z-axis 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 anti-curl 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
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 xy-plane, 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
about upwards pointing vectors), so we need to negate the resulting integral. Thus: F) da 1 1 1 1 1 1 z j + y k ), 1, 1) dt du z y ) dt du u ) dt du 3 14. 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
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 anti-divergence 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
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