Math 6A Practice Problems II Written by Victoria Kala vtkala@math.ucsb.edu SH 64u Office Hours: R : :pm Last updated 5//6 Answers This page contains answers only. Detailed solutions are on the following pages.. 5π. π. a) Not conservative b) onservative f x y + x ln y + 4. a) curlf b) f xy cos z + c) 5. a) 44 b) 88 c) 8 6. a) ln b) 8 5 ) 7. 8. ) 9. 4.. 6 4. 44. 4. 8 + 4 4 5 5. πe 6 e 5) 6.
Detailed Solutions. Evaluate the line integral xyzds where is the curve parametrized by x sin t, y t, z cos t, t π. Solution. Since rt) sin t, t, cos t, r t) cos t,, sin t and r t) cos t) + + sin t) 5. Then xyzds sin t)t) cos t) r t) dt 4 5 5 t cos t + ) 4 sin t π 5π. We used integration by parts to evaluate the integral. t sin t cos tdt 5 t sin tdt. Evaluate the line integral F dr where is given by the vector function rt) ti + sin tj + cos tk, t π and F zi + yj xk. Solution. Frt)) cos t, sin t, t and rt), cos t, sin t. Therefore F dr Frt)) r t)dt cos t, sin t, t, cos t, sin t dt cos t + sin t cos t + t sin t) dt cos t + ) sin t + t sin t dt sin t 4 cos t t cos t + sin t π π. We used integrating by parts to evaluate the integral t sin tdt.. Determine whether or not F is a conservative vector field. If it is, find a function f such that F f. a) Fx, y) e x cos yi + e x sin yj Solution. P e x cos y and Q e x sin y. Then Since Q x P b) Fx, y) ln y + xy )i + P ex sin y, F is not conservative. x y + x ) j y and Q x ex sin y.
Solution. P ln y + xy and Q x y + x y. Then P y + 6xy, and Q x 6xy + y. Since Q x P, F is conservative and there exists an f such that f F. We wish to find a function f such that f x, f y ln y + xy, x y + x y : f x ln y + xy f x ln y + x y + gy) f y x y + x y f x y + x ln y + hx). Therefore f x y + x ln y +. 4. Let Fx, y, z) y cos zi + xy cos zj xy sin zk, be the curve parametrized by rt) t i + sin tj + tk, t π. a) Show that F is conservative. Hint: use curl) Solution. We need to show curlf) : i j k curlf) x z y cos z xy cos z xy sin z i z xy cos z xy sin z j x z y cos z xy sin z + k x y cos z xy cos z i xy sin z) ) xy cos z) j z x xy sin z) ) z y cos z) + k x xy cos z) ) y cos z) xy sin z + xy sin z)i y sin z + y sin z)j + y cos z y cos z)k. Therefore F is conservative. b) Find a function f such that F f. Solution. We need to find an f such that f x, f y, f z y cos z, xy cos z, xy sin z : Therefore f xy cos z +. f x y cos z f xy cos z + gy, z) f y xy cos z f xy cos z + hx, z) f z xy sin z f xy cos z + lx, y) c) Use b) to calculate F dr along the given curve.
Solution. F dr f dr frπ)) fr) fπ,, π) f,, ). 5. a) Estimate the volume of the solid that lies below the surface z x + y and above the rectangle R [, ] [, 4]. Use a Riemann sum with m n and choose the sample points to be the lower right corners. Solution. x and y 4, therefore A x y. The lower righthand corners of each rectangle are, ),, ),, ),, ). Let z fx, y). Then fx, y)da f, ) + f, ) + f, ) + f, )) A + 9 + + ) 44. b) Use the midpoint rule to estimate the volume in a). Solution. We did not change the division of our rectangles, so A. The midpoints of each rectangle are, ),, ),, ),, ). Then ) ) ) )) fx, y)da f, + f, + f, + f, A + + + 8 + + + ) + 8 88. c) alculate the exact volumes of the solid. Solution. 4 x + y )dydx xy + y ) 4 dx 4x + 8 ) dx x + 8 x 8 6. Evaluate the following integrals: 4 x a) y + y ) dydx x Solution. 4 x y + y ) dydx x 4 ) x ln y + y x dx y 4 x ln + ln x ln x 4 ln x ln + x ) dx x ln 4 ln 4
b) Solution. s + t dsdt + t) / t /) dt s + t) / dsdt s + t)/ dt 5 + t)5/ ) 5 t5/ 8 5 ) 7. Evaluate D x + y) da where D is bounded by y x, y x. Solution. The region described is D {x, y) : x, x y x}. D x x + y)da x + y)dydx x 5 x5/ + x 4 x4 4 x5 5 ) xy + y x dx x x / + x ) x x4 dx 8. Evaluate the integral by reversing the order of integration: / arcsin y cos x + cos x dxdy. Solution. The region described in the integral above is equivalent to x π, y sin x: / sin x cos x + cos x dydx / / cos x + cos x y sin x sin x cos x + cos xdx, udu u/ dx let u + cos x ) 9. Evaluate x + y) da where R is the region that lies to the left of the y-axis between the R circles x + y, x + y 4. 5
Solution. The region described in polar coordinates is R {r, θ) : r, π θ π }. / / x + y)da r cos θ + r sin θ)rdrdθ cos θ + sin θ)r drdθ R π/ / π/ cos θ + sin θ) r 7 sin θ cos θ) π/ 4 π/ dθ 7 π/ / π/ cos θ + sin θ)dθ. Verify Green s Theorem for x4 dx + xydy where is the triangular curve consisting of line segments from, ) to, ),, ), to, ), and, ) to, ) traversed in that order. Solution. We need to verify F dr ) Q D x P da. We need to calculate both the left integral and the right integral and show they are equal. For the left integral: We need to break up our triangle into three different sections. Let be the side of the triangle from, ) to, ), be the side of the triangle from, ) to, ), and be the side of the triangle from, ) to, ). Then + +. is parametrized by rt) t,, t. Then F dr F rt)) r t)dt t 4 dt + t dt is parametrized by rt) t, t, t. Then F dr F rt)) r t)dt t) 4 dt) + t)tdt 5 t)5 + t t 5 t 4 dt 5 t5 5. is parametrized by rt), t, t. Then F dr F rt)) r t)dt dt) + t) dt). Therefore F dr 5 + 5 + 6. t) 4 + t t ) dt learly this method is tedious. Now we will apply Green s Theorem. The triangle can be described as D {x, y) : x, y x}. Then Q D x P ) x da x xy) ) x x4 ) dydx ydydx y x x) x) dx dx 6 6. Our two calculations are equal to each other, thus the theorem is verified. 6
. Evaluate y dx + xydy, where is the boundary of the semiannular region D in the upper half plane between the circles x + y and x + y 4. You may assume that is positively oriented.) Solution. D is the region {x, y) : x + y 4} {r, θ) : r, θ π}. Using Green s Theorem, we have y dx + xydy x xy) ) y ) da y y)) da yda D 4 r sin θ rdrdθ sin θdθ D r dr cos θ) π D r. Use a triple integral to find the volume of the solid bounded by the cylinder y x and the planes z, z 4, and y 9. Solution. The solid can be described as E {x, y, z) : x, x x 9, z 4}. Therefore 9 4 9 V dv dzdydx z 4 9 4 dydx 4 y E x x dydx 9 dx x x ) 4 9 x )dx 4 9x x 44. Evaluate E xydv where E is bounded by the parabolic cylinders y x and x y, and the planes z and z x + y. Solution. The solid can be described as E {x, y, z) : x, x y x, z x + y}. Then E xydv x x+y x x y + xy x4 8 + x7/ x7 xydzdydx ) x dx x 4 x8 4 x x x xyz x+y z + x5/ 8 + 4 4 dydx x x6 x7 x ) dx x y + xy )dydx 4. Evaluate E x + xy )dv where E is the solid in the first octant that lies beneath the paraboloid z x y. 7
Solution. The solid can be described in cylindrical coordinates as E {r, θ, z) : r, θ π, z r }. Since x + xy xx + y ) r cos θ r, then E x + xy )dv / r / / r 5 r cos θ rdzdrdθ r 4 r ) cos θdrdθ 5 r7 7 ) / / cos θdθ 5 sin θ π/ 5 r 4 cos θ z r r 4 r 6 ) cos θdrdθ drdθ 5. Evaluate E ez dv where E is enclosed by the paraboloid z + x + y, the cylinder x + y 5, and the xy-plane. Solution. The solid can be described in cylindrical coordinates as E {r, θ, z) : r 5, θ π, z + r }. E e z dv 5 +r 5 π 6 e z rdzdrdθ re +r )drdθ π e u du r 5 ) 5 ) π e u 6 5 re z +r z drdθ re +r dr πe 6 e 5) ) rdr, let u + r 6. Evaluate E xyzdv where E lies between the spheres ρ and ρ 4 and the cone φ π. Solution. The solid can be described in polar coordinates as E {ρ, θ, φ) : ρ 4, θ π, φ π }. Then E xyzdv 4 / 4 6 ρ6 4 ρ 5 dρ sin θ ρ cos θ sin φ)ρ sin θ sin φ)ρ cos φ) ρ sin φdρdθdφ sin θ cos θdθ π 4 sin4 φ / π/ sin φ cos φdφ 8