Suppose $f(x,y,z)=g\left(\sqrt{x^2+y^2+z^2}\right)$, where $g$ is a function of one variable such that $g(2)=-5$. Evaluate $$\iint_S f ~dS,$$where $S$ is the sphere $x^2+y^2+z^2=4$.
Now, I reasoned as follows. If $g(2)=-5$ then when $\sqrt{x^2+y^2+z^2}=2 \implies x^2+y^2+z^2=2$ the function $f$ takes the value of $-5$. Hence, since $S$ is the sphere of radius $4$ centred at the origin, we have \begin{align*}\iint_S f ~dS&=-5\iint_\Omega \left\|\frac{\partial\mathbf r}{\partial\phi }\times\frac{\partial\mathbf r}{\partial\theta } \right\|~dA,\end{align*}where $\mathbf{r}(\phi,\theta)$ is a parametrisation of $S$ using spherical coordinates: $$\mathbf{r}(\phi,\theta)=\left(2\sin\phi\cos \theta,2\sin\phi\sin\theta,2\cos\phi\right), ~(\phi,\theta)\in[0,\pi]\times[0,2\pi].$$
Hence \begin{align*}\iint_S f ~dS&=-5\int\limits_0^{2\pi}\int\limits_0^\pi 4\sin\phi~d\phi d\theta\\&=-80\pi.\end{align*}
Can somebody please verify that my reasoning (especially in the first steps) is correct? The minus sign confuses me for some reason; can surface integrals be negative?
Best Answer
The answer is correct and, actually, no integration is required: $$\iint_S f\,dS = \iint_S (-5)\,dS = (-5)\text{area}(S) = (-5) 4\pi 2^2.$$