I've been asked to solve the following problem.
Find the value of the integral $\int \mathbf A \cdot d\mathbf a$ , where $\mathbf A = (x^2+y^2+z^2)(xi+yj+zk)$ and the surface $S$ is defined by the sphere $R^2=x^2+y^2+z^2$. Do the integral directly and by Gauss's theorem.
Doing the problem with Gauss's formula is easy enough, but computing it directly is throwing me for a loop. I've tried using the formula below to compute it, but ended up with a very messy final answer, which is causing me to doubt my work. $$\iint \mathbf F \cdot d\mathbf S = \iint \mathbf F \cdot \mathbf n dS$$
I used the this website as a guide to solving it, but again, I am not confident in my final answer.
How would you approach this problem and if someone has the time, what is the final answer?
Thank you.
Let $B$ be the ball centred at the origin with radius $R$ and let $\mathbf F = (x^2+y^2+z^2)(x\mathbf i+y \mathbf j+z \mathbf k)$.
By direct computation, since the area of the sphere $\partial B$ is $4\pi R^2$, we get $$\iint_{\partial B} \mathbf F \cdot d\mathbf S= \iint_{\partial B} \mathbf F \cdot \mathbf n\ dS=R^3\iint_{\partial B} \ dS=R^3\cdot 4\pi R^2=4\pi R^5$$ because for any point $(x,y,z)\in\partial B$ $$\mathbf F \cdot \mathbf n= (x^2+y^2+z^2)(x\mathbf i+y \mathbf j+z \mathbf k)\cdot(x \mathbf i+y \mathbf j+z \mathbf k)/R=(x^2+y^2+z^2)^2/R=R^3$$ where $\mathbf n=(x \mathbf i+y \mathbf j+z \mathbf k)/R$.
On the other hand, by Gauss's theorem, $$\iint_{\partial B} \mathbf F \cdot d\mathbf S=\iiint_B\mbox{div}(\mathbf F)\ dV.$$ Now $$\mbox{div}(\mathbf F)=(3x^2+y^2+z^2)+(x^2+3y^2+z^2)+(x^2+y^2+3z^2) =5(x^2+y^2+z^2).$$ By using spherical coordinates we get $$\iiint_B z^2 dV=\int_{\theta=0}^{2\pi}\int_{r=0}^R\int_{\phi=0}^{\pi} (r\cos(\phi))^2 (r^2\sin(\phi))\ d\phi dr d\theta\\=2\pi\left[\frac{r^5}{5}\right]_0^R \left[-\frac{\cos^3(\phi)}{3}\right]_0^{\pi}=\frac{4\pi R^5}{15}.$$ Moreover, by symmetry, $\iiint_B x^2 dV$ and $\iiint_B y^2 dV$ yield the same result. Thus $$\iiint_B\mbox{div}(\mathbf F)\ dV=3\cdot 5\cdot \frac{4\pi R^5}{15}=4\pi R^5.$$