Final Exam Quick Solution Sheet

December 22, 1999

1. (\nabla f)(x, y, z) = ( 2 x y, x^{2} - 3 y^{2} z^{2}, - 2 y^{3} z )

2. (curl F)(x, y, z) = ( -3 z^{2}, -3 x^{2}, -3 y^{2})

3. INT[_{0}^{2} x^{2} d x INT[_{0}^{4} y d y ]] = {64}/{3}

4. - 12 x + 2 y + 3 z = 24

5. r'(t) = (-2 sin t, 2 cos t , 1)

   ||r'(t)|| = SQRT{5}

   L = ({Pi}/{2}) SQRT{5}

6. F = \nabla f where f(x, y, z) = x^{2} + x y - y z + z

   INT[_{C} F] = f(r(2)) - f(r(0)) = f(r(2)) = 76

7. 22 x + 27 y + 21 z = 101

8. INT[INT[_{\partial B} F ]] = INT[INT[INT[_{B} div F ]]] = 0 since div F = 0

9. 1. (- cos t , 1, e^{t} - 1)

      ( sin t , 0, e^{t})

      SQRT{2}

      {1}/{2}

      (0, 0, 1)

   2. (0, 0, 15/4) Note: V = {125 pi}/{3}

10. This F is a curl. (See problem 2.) So Stoke's Theorem can be used to simplify and to replace the given surface integral by an ordinary double integral over the disk.

    INT[INT[_{Delta} {(z^{2}, x^{2}, y^{2}) . (0, 0, 1)} d x d y ]] = {81 pi}/{4}