10C (Vector Calculus)

The divergence theorem states that a continuously differentiable vector field on a bounded region with piecewise smooth boundary satisfies

\int_{V} \nabla \cdot u d V = \int_{\partial V} u \cdot n d S,

with the unit normal pointing out of

V

. Apply it separately to

u = f e_{i}

, where

e_{i}

is each constant Cartesian basis vector. Since

\nabla \cdot (f e_{i}) = \partial_{i} f

and

(f e_{i}) \cdot n = f n_{i}

, the three component identities combine to give

\int_{V} \nabla f d V = \int_{\partial V} f d S .

For the specified scalar field,

\nabla f = (z, 0, x)

. The

x

term integrates to zero by reflection symmetry. At height

z

, the region has a disk cross-section of area

π (a^{2} - cz)

, and

0 \leq z \leq a^{2} / c

. Therefore the volume integral is

\int_{V} \nabla f d V = e_{x} π \int_{0}^{a^{2} / c} z (a^{2} - cz) d z = \frac{π a ^{6}}{6 c ^{2}} e_{x} .

To check the surface integral independently, split the boundary into the base disk and the curved graph

z = g (x, y) = (a^{2} - x^{2} - y^{2}) / c

. On the base

f = 0

, so its contribution vanishes. The region lies below the curved graph, whose outward vector surface element of a graph is

d S = (- g_{x}, - g_{y}, 1) d x d y = (\frac{2 x}{c}, \frac{2 y}{c}, 1) d x d y .

Over the disk

x^{2} + y^{2} \leq a^{2}

, the

y

and

z

components of

f d S

are odd in

x

and integrate to zero. Its

x

component gives, in polar coordinates,

\int_{\partial V} f d S = \frac{2 e _{x}}{c ^{2}} \int_{x^{2} + y^{2} \leq a^{2}} x^{2} (a^{2} - x^{2} - y^{2}) d x d y = \frac{2 e _{x}}{c ^{2}} \int_{0}^{2 π} cos^{2} ϕ d ϕ \int_{0}^{a} s^{3} (a^{2} - s^{2}) d s = \frac{π a ^{6}}{6 c ^{2}} e_{x} .

Thus the two explicitly evaluated integrals agree, with the outward orientation included.

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