Compute $\unicode{x222F}_{\partial K} |\nabla f|dS$ where $f(x,y,z)=x^2+4y^2+9z^2 , K=\{(x,y,z);f(x,y,z)\leq 1\}$
I came across this question when studying for an upcoming exam, but it had no answer attached, so I would like to know if you agree with my solution.
$\textbf{Solution}$: Assuming we are calculating the outwards flux, we note that $\hat{N}=\frac{\nabla f}{|\nabla f|}$ for surface $K$. Then, suppose $\bar{F}=\nabla f$, which would give us that $\bar{F} \cdot \hat{N}=|\nabla f|$. By the divergence theorem, we know that $$\iint_{\partial K}\bar{F} \cdot \hat{N} dS=\iiint_{K}\nabla \cdot F dV$$ where the former integral is a closed integral. Now, $\nabla f = 2x\hat{i}+8y\hat{j}+18z\hat{k}$, so $\nabla \cdot F= \nabla \cdot \nabla f= 28$. Thus, $$\iint_{\partial K} |\nabla f| dS=28\iiint_{x^2+4y^2+9z^2 \leq 1}dV$$ Letting $x=u,y=\frac{1/2}v,z=\frac{1/3}w$, we find that $dxdydz=\frac{1}{6}dudvdw$ and subsequently that $$\iint_{\partial K} |\nabla f| dS=\frac{28}{6}\iiint_{u^2+v^2+w^2 \leq 1}dV$$ And finally, $$\iint_{\partial K} |\nabla f| dS=\frac{28}{6}(\frac{4}{3}\pi)=\frac{56\pi}{9}$$ Does my solution seem correct?
Looks good! I get the same result by directly attacking the surface integral. Parameterize $\partial K$ in spherical coordinates by
$$\hat r(u,v) = \cos u \sin v \, \hat\imath + \frac12 \sin u \sin v \, \hat\jmath + \frac13 \sin v \, \hat k$$
with $(u,v)\in[0,2\pi]\times[0,\pi]$ so that the normal to $\partial K$ (note that orientation doesn't matter here) and its norm are, respectively,
$$\begin{align*} \hat n &= \frac{\partial\hat r}{\partial v} \times \frac{\partial\hat r}{\partial u} = \frac16 \cos u \sin^2v \,\hat\imath + \frac13 \sin u \sin^2v \, \hat\jmath + \frac12 \sin v \cos v \, \hat k \\ \left\|\hat n\right\| &= \sqrt{\frac1{36} \cos^2u \sin^4v + \frac19 \sin^2u \sin^4v + \frac14 \sin^2v \cos^2v} \end{align*}$$
The integrand evaluates to
$$\begin{align*} \left\|\nabla f\left(\hat r(u,v)\right)\right\| &= \left\|2\cos u\sin v\,\hat\imath + 4\sin u \sin v\,\hat\jmath + 6\sin v\,\hat k\right\| \\ &= \sqrt{4\cos^2u\sin^2v + 16\sin^2u\sin^2v+36\sin^2v}\end{align*}$$
Combining the above expressions and simplifying reduces the surface integral to an elementary double integral whose value agrees with your result,
$$\iint_{\partial K} \left\|\nabla f\right\| \, dS = \int_0^\pi \int_0^{2\pi} \left(3\sin v + \left(\sin^2u - \frac83\right) \sin^3v\right) \, du \, dv = \boxed{\frac{56\pi}9}$$