9B (Vector Calculus)

The surface is an annular band of the downward-opening paraboloid, bounded by the circles

r = 1, z = 4

and

r = 2, z = 1

. Choose the upward orientation.

The curl of

A = (3 y, - x z, y z^{2})

\nabla \times A = (z^{2} + x, 0, - z - 3) .

For the graph

z = 5 - x^{2} - y^{2}

, the upward vector surface integral element is

d S = (2 x, 2 y, 1) d x d y .

Over the annulus

1 < r < 2

, the odd term

2 x z^{2}

integrates to zero. Therefore

I = \iint_{1 < r < 2} (2 x^{2} - z - 3) d x d y = \int_{1}^{2} \int_{0}^{2 π} (2 r^{2} cos^{2} θ + r^{2} - 8) r d θ d r = \int_{1}^{2} (4 π r^{3} - 16 π r) d r = - 9 π .

b

I = \oint_{\partial S} A \cdot d r .

For the upward orientation, the outer circle

r = 2, z = 1

is counterclockwise from above and the inner circle

r = 1, z = 4

is clockwise. On a counterclockwise circle of radius

R

z

\oint A \cdot d r = - R^{2} \int_{0}^{2 π} (3 sin^{2} θ + z cos^{2} θ) d θ = - π R^{2} (3 + z) .

The outer circle contributes

- 16 π

. The inner circle would contribute

- 7 π

counterclockwise and hence contributes

+ 7 π

with its induced clockwise orientation. Thus

I = - 16 π + 7 π = - 9 π,

agreeing with the direct calculation.

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