10A (Vector Calculus)

Solution

S

\int_{S} (\nabla \times B) \cdot n d S = \oint_{\partial S} B \cdot d r,

where the boundary orientation is induced by

n

through the right-hand rule. Choose the upward orientation of the elliptic paraboloid. Its band has cylindrical radius

1/2 \leq r \leq 1

, and a parametrization is

X (r, φ) = (r cos φ, r sin φ, r^{2}), 0 \leq φ < 2 π .

d S = (X_{r} \times X_{φ}) d r d φ = (- 2 r^{2} cos φ, - 2 r^{2} sin φ, r) d r d φ,

d S = r 1 + 4 r^{2} d r d φ .

The requested sketch shows the open band, not a capped solid.

For the specified vector field, direct differentiation gives

\nabla \times B = (0, 0, 3 x^{2} + 3 y^{2})

. Hence

I = \int_{0}^{2 π} \int_{1/2}^{1} 3 r^{3} d r d φ = \frac{45 π}{32} .

To check this by Stokes theorem, parametrize a counterclockwise circle of radius

ρ

ρ^{2}

r_{ρ} (φ) = (ρ cos φ, ρ sin φ, ρ^{2})

. Along it

d z = 0

, and

B \cdot d r_{ρ} = ρ^{4} (sin^{4} φ + cos^{4} φ) d φ .

Using

\int_{0}^{2 π} sin^{4} φ d φ = \int_{0}^{2 π} cos^{4} φ d φ = 3 π /4

gives

3 π ρ^{4} /2

. The induced outer circle is counterclockwise as viewed from above, and the inner circle is clockwise. Therefore

I = \oint_{C_{1}} B \cdot d r - \oint_{C_{1/2}} B \cdot d r = \frac{3 π}{2} (1 - \frac{1}{16}) = \frac{45 π}{32} .

This verifies the direct surface integral and the Stokes flux through an annular paraboloid. Reversing the chosen surface orientation reverses the final sign.

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