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 Incoming links: Surface integral

Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 10B / i / Solution Created 2026-09-24 Updated 2026-09-25

Parametrize the upward-oriented paraboloid by

X (r, θ) = (r cos θ, r sin θ, 1 - r^{2}), 0 \leq r \leq 1, 0 \leq θ \leq 2 π .

Then

X_{r} \times X_{θ} = (2 r^{2} cos θ, 2 r^{2} sin θ, r) .

On the surface,

F = (r^{2} cos^{2} θ, - r^{2} sin^{2} θ, (1 - r^{2})^{4}) .

The cubic trigonometric terms integrate to zero over

0 \leq θ \leq 2 π

, so the surface integral is

\int_{S} F \cdot d S = 2 π \int_{0}^{1} r (1 - r^{2})^{4} d r = \frac{π}{5} .

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Stokes flux through an annular paraboloid 2026-10-07

For an upward-oriented band of the circular elliptic paraboloid between radii

r_{0}

and

R

, the vector field

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

has vertical curl

3 (x^{2} + y^{2})

. Its surface integral is

3 π (R^{4} - r_{0}^{4}) /2

. Stokes theorem gives the same value from the counterclockwise outer circle minus the counterclockwise inner circle. Equivalently the induced inner boundary direction is clockwise. Forgetting the inner boundary or its reversed orientation changes the answer.

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Vector surface element of a graph 2026-10-06

z = g (x, y)

has upward-oriented vector area element

d S = (- g_{x}, - g_{y}, 1) d x d y

and scalar area element

d S = 1 + g_{x}^{2} + g_{y}^{2} d x d y

. These follow from the cross product of the parametrization derivatives

(1, 0, g_{x})

and

(0, 1, g_{y})

. The vector element is used in flux surface integrals and Stokes theorem; reversing orientation changes its sign.

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