Jim Lambers MAT 280 Summer Semester Practice Final Exam Solution. dy + xz dz = x(t)y(t) dt. t 3 (4t 3 ) + e t2 (2t) + t 7 (3t 2 ) dt

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1 Jim Lambers MAT 28 ummer emester Practice Final Exam olution 1. Evaluate the line integral xy dx + e y dy + xz dz, where is given by r(t) t 4, t 2, t, t 1. olution From r (t) 4t, 2t, t 2, we obtain 1 xy dx + e y dx dy + xz dz x(t)y(t) dt 2. how that the vector field ey(t) dy dt + x(t)z(z)dz dt dt t (4t ) + e t2 (2t) + t 7 (t 2 ) dt 4t 6 + 2te t2 + t 9 dt ( ) 4t et2 + t1 1 1 ( e + ) 1. 1 F(x, y, z) e y, xe y + e z, ye z is conservative, and use this fact to evaluate F dr where is the line segment from (, 2, ) to (4,, ). olution A vector field F P, Q, R is conservative if From R y Q z, P z R x, Q x P y. R y e z Q z, P z R x, Q x e y P y, we find that F is in fact conservative. To evaluate the line integral efficiently, we need to find a function f such that f F. To that end, we obtain f(x, y, z) P dx xe y + g(y, z). The requirement that f y Q yields the equation olving the equation for g y yields g y (y, z) Q(x, y, z) (xe y ) y e z. g(y, z) ye z + h(z). The requirement that f z R yields the equation h (z) R(x, y, z) (xe y + ye z ) z.

2 It follows that h(z) K where K is an arbitrary constant, and therefore f(x, y, z) xe y + ye z + K. From the Fundamental Theorem of Line Integrals, we obtain F dr f dr f(4,, ) f(, 2, ) (4e + e ) (e 2 + 2e ) 2.. Use Green s Theorem to evaluate x 2 y dx xy 2 dy, where is the circle x 2 + y 2 4 with counterclockwise orientation. olution Let D { (x, y) x 2 + y 2 4 } be the interior of. Then, by Green s Theorem, x 2 y dx xy 2 dy ( xy 2 ) x (x 2 y) y da y 2 x 2 da. onverting to polar coordinates, we obtain x 2 y dx xy 2 dy D 2π 2 2π 2π 8π. 2 dθ ( r4 4 D ( r 2 )r dr dθ ) 2 r dr 4. If f(x, y, z) and g(x, y, z) are twice differentiable functions, show that where 2 f ( f). olution We have 2 (fg) ( (fg)) 2 (fg) f 2 g + g 2 f + 2 f g, (fg) x, (fg) y, (fg) z f x g + fg x, f y g + fg y, f z g + fg z (f x g + fg x ) x + (f y g + fg y ) y + (f z g + fg z ) z f xx g + 2f x g x + fg xx + f yy g + 2f y g y + fg yy + f zz g + 2f z g z + fg zz f(g xx + g yy + g zz ) + g(f xx + f yy + f zz ) + 2(f x g x + f y g y + f z g z ) f ( g) + g ( g) + 2 f g f 2 g + g 2 f + 2 f g.

3 5. Evaluate the surface integral (x 2 z + y 2 z) d, where is the part of the plane z 4 + x + y that lies inside the cylinder x 2 + y 2 4. olution We use the parametric equations for which x u cos v, y u sin v, z 4 + u cos v + u sin v, u 2, v 2π, r u cos v, sin v, cos v + sin v, r v u sin v, u cos v, u sin v + u cos v, This yields (x 2 z + y 2 z) d r u r v u, u, u, 2π 2 2π 2 2π 2π r u r v u. [(u cos v) 2 + (u sin v) 2 ](4 + u cos v + u sin v) u du dv 4u + u 4 cos v + u 4 sin v du dv [u 4 + u5 (cos v + sin v) (cos v + sin v) dv 5 ] 2 dv 16 (2π) π. (sin v cos v) 2π 6. Evaluate the surface integral F d, where F(x, y, z) xz, 2y, x and is the sphere x 2 + y 2 + z 2 4 with outward orientation. olution We use spherical coordinates x 2 sin φ cos θ, y 2 sin φ sin θ, z 2 cos φ, θ 2π, φ π, which yields r φ 2 cos φ cos θ, 2 cos φ sin θ, 2 sin φ, r θ 2 sin φ sin θ, 2 sin φ cos θ,, r φ r θ 4 sin φ sin φ cos θ, sin φ sin θ, cos φ,

4 which has outward orientation since 4 sin φ for φ π. We then have 2π π F d 2 sin φ 2 cos θ cos φ, 2 sin θ, cos θ 4 sin φ sin φ cos θ, sin φ sin θ, cos φ dφ dθ π π π 2π 2π 2π sin 2 φ(2 cos 2 θ sin φ cos φ 2 sin 2 θ sin φ + cos θ cos φ) dφ dθ cos 2 θ dθ cos θ dθ π π 1 + cos 2θ dθ 8 1 cos 2θ dθ 16π sin4 π φ π ( 16π cos φ cos φ ( 16π ) 64π. 2π π sin φ cos φ dφ 16 sin 2 θ dθ sin φ dφ + sin 2 φ cos φ dφ π 1 1 ) π u du (u sin φ, du cos φ dφ) (1 cos 2 φ) sin φ dφ + 24 sin θ 2π π (1 v 2 ) dv (v cos φ, dv sin φ dφ) sin 2 φ cos φ dφ 7. Use tokes Theorem to evaluate F dr, where F(x, y, z) xy, yz, xz and is the triangle with vertices (1,, ), (, 1, ) and (,, 1), oriented counterclockwise as viewed from above. Hint: to obtain an equation for the surface enclosed by, compute the equation of a plane containing the vertices. olution Using the given vertices, it can be determined that is contained within the plane x + y + z 1. We therefore describe using the parametric equations which yields and the normal vector x u, y v, z 1 u v, u 1, v 1 u, r u 1,, 1, r v, 1, 1, r u r v 1, 1, 1, which is consistent with the counterclockwise orientation of (that is, when traversing such that this normal vector, which points upward, is visible, then the region is on the left. Applying tokes Theorem yields 1 1 u F dr curl F(u, v) (r u r v ) dv du. From curl F R y Q z, P z R x, Q x P y (xz) y (yz) z, (xy) z (xz) x, (yz) x (xy) y y, z, x.

5 We then have and thus We conclude that curl F (r u r v ) y, z, x 1, 1, 1 (x + y + z), curl F(u, v) (r u r v ) (u + v + 1 u v) 1. F dr 1 1 u 1 dv du A(D), where D is the triangle { (u, v) u 1, v 1 u }. This triangle has base and height 1, which yields F dr Use the Divergence Theorem to evaluate the surface integral F d, where F(x, y, z) x, y, z and is the surface of the solid E bounded by the cylinder x 2 + y 2 1 and the planes z and z 2. olution Using cylindrical coordinates, we obtain F d div F dv E [(x ) x + (y ) y + (z ) z ] dv E (x 2 + y 2 + z 2 ) dv 9. (Bonus) Let and let 2π E 1 2 2π 1 2π 1 dθ ( ) r 4 6π 2 + 4r2 1 11π. (r 2 + z 2 )r dz dr dθ ) (r z + r z 2 dr dθ 2r + 8 r dr f(x, y, z) x 2 y z 4, g(x, y, z) x 4 + y + z 2, ω f dx + g dy, η f dx dz + g dy dz be a 1-form and a 2-form, respectively. ompute ω η and dω. olution We have ω η (f dx + g dy)(f dx dz + g dy dz) f 2 dx dx dz + fg dy dx dz + fg dx dy dz + g 2 dy dy dz fg dx dy dz + fg dx dy dz +,

6 and dω df dx + dg dy (f x dx + f y dy + f z dz) dx + (g x dx + g y dy + g z dz) dy f x dx dx + f y dy dx + f z dz dx + g x dx dy + g y dy dy + g z dz dy (g x f y ) dx dy + f z dz dx g z dy dz (4x x 2 y 2 z 4 ) dx dy + 4x 2 y z dz dx 2z dy dz.