1. (16 points) Write but do not evaluate the following integrals:

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1 MATH xam # Solutions. (6 points) Write but do not evaluate the following integrals: (a) (6 points) A clindrical integral to calculate the volume of the solid which lies in the first octant (where x,, and z coordinates are all positive) under the paraboloid z x and above the cone z x +. The surface bounding this solid above can be written as z (x + ) r ; the lower bound is clearl z r. These two solids intersect when r r, which occurs when r + r, which is satisfied at the nonsensical value r and the eminentl sensible value r ; thus, this surface s outermost edge is r. The restriction to the first octant requires that z, x, and. The first of these restrictions is moot: the previous restriction z r ensures that z is non-negative; however, the restrictions on x and constrain θ to those values whose cosines and sines are both positive, which is to sa, the first-quadrant section θ π. Our integral is thus: π/ r dv rdzdrdθ (b) (5 points) A polar integral to calculate e x da, where is the region given b x + 4 with and x. The region in question consists of those values where r, sin θ, and cos θ sin θ. The former constraint on θ we know to cover the first and second quadrants, which are the range [, π]; however, for θ < π, θ s cosine exceeds its sine, so x > ; we must exclude 4 this region as per the second theta-constraint, so π θ π. Translating the integrand 4 and multipling b the integrating factor r, we get: e x da π π/4 r re r drdθ (c) (5 points) A spherical integral to calculate x + dv where is the hollow hemispherical shell given b x + + z 9 with. As seen above, when θ π; in addition, since x + + z ρ, the first constraint is that ρ. φ is, in this case, unrestricted. Finall, the integrand x + expands to ρ sin φ(cos θ + sin θ) ρ sin φ, and so: x + dv π π (ρ sin φ)(ρ sin φ)dρdφdθ. (6 points) alculate the following integrals, using whatever approach ou find most effective: (a) (6 points) 5dV where is the solid in the first octant bounded b the surfaces x,, x, z x, and z 4. The surfaces z x and z 4 meet when x 4; in the first octant, this guarantees that the solid lies within x. The coordinate is bounded b and x, so x; finall, since the solid lies between z x and z 4, x z 4, and so we have an Page of 5 Monda, November,

2 MATH xam # Solutions integral: 5dV x 4 x x 5dzddx 5z ] z4 zx ddx 5 ] x x dx x x5 ] 8 x 6 5x ddx x 5 x4 dx (b) (6 points) x da where is the region bounded b the curves x and x. These curves intersect when x x, or where x, or at the points x ±, so the region is given b x, and bounded above and below b x and x respectivel. Thus, our integral is: x da x x x ddx x ] x x dx x 4 x4 4x + 4 x ] x x (4 4x )dx ( ) ( ) 4( + ) ( + ) (c) (4 points) x + da where is the rectangle with corners (, ), (, ), (, ), and (, ). Since this is an integral over a rectangular region given b x and, the integral can be set up as a simple iterated integral with constant limits: x + da x + ddx x + ] dx 4x + dx ] x + x (4 9) + (5). (6 points) Using the transformations x u v and u + 4v, evaluate x + da over the region bounded b 4x + 8, 4x + 7, x, and x 9. Note that these four boundaries can be re-expressed easil in terms of u and v. 4x + 4(u v) + (u + 4v) 9u, so the inequalit 8 4x + 7 becomes u with ease; likewise x (u + 4v) (u v) 9v, so x 9 becomes v. Now, we need onl re-express the integrand and calculate the Jacobian. The integrand is x + (u v) + (u + 4v) u + v; the Jacobian is calculated here: (x, ) (u, v) x u u x v v 4 9 Page of 5 Monda, November,

3 MATH xam # Solutions so our integral can be transformed as follows: x + da (u + v) 9dvdu 7uv u + 7 u v ] ] v du v 7u + 7 du (9 + 4 ) 4. (8 points) Identif each of the following vector fields as either conservative or nonconservative; for each that is conservative, find a potential function: 8 F (x, ) (4x + )i + (4 + )j. We calculate (4x +), and (4+), so this vector field is nonconservative, x since these results are nonequal. G(x, ) x +, x. We calculate ( x x + ), and x ( x ) x, so this vector field is conservative. Using the partial integrals of each component, we find that the potential function g(x, ) is given b and g(x, ) g(x, ) x + dx x + x + () x d + x + (x) Although these appear different, the term x in the first integral is represented in the second within the junk term (x), and likewise the term in the second integral is subsumed into the first integral s junk term (). Thus, a potential function that matches both descriptions is g(x, ) x + x +. H(x, ) ln e x, 7 sin + x. We calculate (ln ex ), and (7 sin + x), so this vector field is conservative. x Using the partial integrals of each component, we find that the potential function h(x, ) is given b h(x, ) ln e x dx x ln e x + () and h(x, ) 7 sin + x d 7 cos + x ln + (x) Although these appear different, the term e x in the first integral is represented in the second within the junk term (x), and likewise the term 7 cos in the second integral is subsumed into the first integral s junk term (). Thus, a potential function that matches both descriptions is h(x, ) x ln 7 cos e x. 5. (4 points) alculate the following path integrals. Page of 5 Monda, November,

4 MATH xam # Solutions (a) (5 points) x ds where is the line segment from (, 4) to (, ). Let us parameterize this segment with x t, 4 t, for t. Then this integral is: x ds x(t) x (t) + (t) dt t 9t + ( ) dt ] (b) (5 points) F dr, where F (x,, z) 4 + z, x z, z and is the curve given b x t, t, and z t from (,, ) to (, 4, ). Since a parameterization is alread given (and the curve in question is from t to t, we ma evaluate this directl: F dr 4(t) + z(t), x(t) z(t), z(t) x (t), (t), z (t) dt 4t + t, t, t, t, dt 8t + tdt 8 t + t ] (c) (4 points) x d where is the curve x from (, ) to (4, 5). Using the parameterization x t, t on t 4, this integral becomes: 4 x d [x(t) (t)] (t)dt 4 4t 6t + 6tdt 4 t ] 4 t4 + t (5 point bonus) On the back of this sheet, identif the shape of the solid whose volume is described b the integral π π/4 csc φ ρ sin φdρdφdθ, and calculate its volume without taking π/4 an integral. This integral is obviousl dv, or the volume of, for given b θ π, π 4 φ π 4, and ρ csc φ. The shape has a full range of θ-values and no dependencies of the other parameters on θ, so it can be expected to be radiall smmetric about the z-axis. onsidering the φ-bounds, which are constant, we can expect this solid to be bounded between two cones. When φ π, we know that z is positive (since cos π is positive, and x z Page 4 of 5 Monda, November,

5 MATH xam # Solutions tan π 4, so the half-cone z x + describes the surface φ π 4 ; likewise, when φ π 4, z x +, so two bounds on the solid are the cones z x + and z x +. Now we consider the ρ-bound ρ csc φ. Multipling both sides b sin φ (which is guaranteed to be positive), we have ρ sin φ ; note that ρ sin φ x +, so this bound is the same as saing that our solid lies inside the clinder x +. So, our solid is a clinder of radius bounded above and below b right cones whose radius equals their height and which meet at the origin. We can thus see that this figure is a clinder of height 6, with two cones of height and radius carved out. We ma calculate this volume to be: 6(π ) (π ) 6π Page 5 of 5 Monda, November,