Closed-Surface Flux Integral via the Divergence Theorem (Paraboloid Capped by a Disk): 04-BS-1, December 2015, Q5

Question 5: Closed-Surface Flux Integral via the Divergence Theorem (Paraboloid Capped by a Disk) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Solid region $V$: bounded above by the paraboloid $z=4-x^2-y^2$ and below by the plane $z=0$ (a dome sitting on the disk $x^2+y^2\le4$, where the paraboloid meets $z=0$). $S=\partial V$ is the entire closed boundary. $\mathbf F=(x^2,\,-2xy,\,x^2z)$.

Find. $\displaystyle\iint_S\mathbf F\cdot\mathbf n\,dA$.

$V$: the dome under the paraboloid $z=4-x^2-y^2$, capped below by the disk $x^2+y^2\le4$ at $z=0$; $S=\partial V$ is the full closed boundary.

Approach. $S$ is closed (bounds the solid $V$), so apply the divergence theorem, $\iint_S\mathbf F\cdot\mathbf n\,dA=\iiint_V\nabla\cdot\mathbf F\,dV$, and evaluate the resulting volume integral in cylindrical coordinates.

Compute the divergence. $$\nabla\cdot\mathbf F=\frac{\partial(x^2)}{\partial x}+\frac{\partial(-2xy)}{\partial y}+\frac{\partial(x^2z)}{\partial z}=2x-2x+x^2=x^2.$$
Set up the volume integral in cylindrical coordinates. With $x=r\cos\theta$, the region is $0\le\theta\le2\pi$, $0\le r\le2$, $0\le z\le4-r^2$, and $dV=r\,dz\,dr\,d\theta$: $$\iiint_V x^2\,dV=\int_0^{2\pi}\cos^2\theta\,d\theta\int_0^2 r^3(4-r^2)\,dr.$$
Evaluate the two 1-D integrals. $$\int_0^{2\pi}\cos^2\theta\,d\theta=\pi,\qquad \int_0^2 r^3(4-r^2)\,dr=\int_0^2(4r^3-r^5)\,dr=\left[r^4-\frac{r^6}{6}\right]_0^2=16-\frac{64}{6}=\frac{16}{3}.$$
Multiply. $$\iiint_V x^2\,dV=\pi\cdot\frac{16}{3}=\frac{16\pi}{3}.$$

$$\iint_S\mathbf F\cdot\mathbf n\,dA=\boxed{\dfrac{16\pi}{3}}\approx16.76$$

Quantity	Result
$\nabla\cdot\mathbf F$	$x^2$
$\int_0^{2\pi}\cos^2\theta\,d\theta$	$\pi$
$\int_0^2 r^3(4-r^2)\,dr$	$16/3$
Total flux	$16\pi/3\approx16.76$

Concept & Theory

Topic. Divergence (Gauss's) theorem for flux through a closed surface, evaluated in cylindrical coordinates.

Key relations. $\displaystyle\iint_{\partial V}\mathbf F\cdot\mathbf n\,dA=\iiint_V(\nabla\cdot\mathbf F)\,dV$ for any closed surface $\partial V$ bounding a solid $V$ with $\mathbf F$ continuously differentiable on $V$. For a paraboloid-capped-by-a-plane region, cylindrical coordinates ($x=r\cos\theta$) turn the $z$-limits into the simple bound $0\le z\le4-r^2$.

Why this works. $S$ is explicitly "the boundary of the region enclosed by" a paraboloid and a plane — a full closed surface — which is exactly the divergence theorem's hypothesis. This avoids separately parametrizing the curved paraboloid cap and the flat disk (and combining two outward-normal orientations) in favour of one volume integral. The divergence $x^2=r^2\cos^2\theta$ separates cleanly into a $\theta$-integral and an $r$-integral because the region's $r,z$ bounds don't depend on $\theta$.

Common pitfall. Forgetting the extra factor of $r$ from $dV=r\,dz\,dr\,d\theta$ in cylindrical coordinates; also attempting to evaluate the paraboloid and disk pieces of $S$ separately (two flux integrals with matched orientations) instead of recognizing the closed-surface shortcut.

Source. Stewart, Calculus: Early Transcendentals, Ch. 16 (§16.9 The Divergence Theorem); Kreyszig, Advanced Engineering Mathematics, Ch. 10 (§10.7 Gauss's Divergence Theorem).

Question 5 of 8: Closed-Surface Flux Integral via the Divergence Theorem (Paraboloid Capped by a Disk)

Question 5: Closed-Surface Flux Integral via the Divergence Theorem (Paraboloid Capped by a Disk) (20 marks)