# Using Divergence Theorem to evaluate the integral

Find the value of $$\iint_{\Sigma} <x, y^3, -z>. d\vec{S}$$ where $\Sigma$ is the sphere $x^2 + y^2 + z^2 = 1$ oriented outward by using the divergence theorem.

So I calculate $div\vec{F} = 3y^2$ and then I convert $x, y, z$ into $x = p\sin \phi \cos \theta, y = p\sin \phi \sin \theta, z = p\cos \phi$ but then I got stuck from that point.

• Did you try and write down the volume integral: $0 \leq p \leq 1, 0 \leq \phi \leq \pi, 0 \leq \theta \leq 2\pi$ – Simon S Dec 17 '15 at 8:10
• The divergence is $1+3y^2-1$. – Emilio Novati Dec 17 '15 at 8:17

$$\iiint_V 3y^2 \ dV = \int_0^1 \int_0^{2\pi} \int_0^\pi 3(p\sin\phi \sin\theta)^2 . p^2 \sin\phi \ dp \ d\theta \ d\phi = \ \cdots$$
Since the volume of integration is a rectangular box in $(p,\theta, \phi)$ coordinates, you can separate out the integrating variables to make it easy to evaluate the overall integral.
• Got it thanks. I was stuck on evaluate $\sin^3 \phi \; d\phi$ so I thought there is something wrong in my work, but I figure it out now. – user298251 Dec 17 '15 at 8:20
You need to evaluate: $$\int_0^1 3 p^4 \ dp\int_0^{2\pi} \sin^2\theta \ d\theta \int_0^\pi \sin^3\phi \ d\phi$$
$$\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^1 \left(\rho^2 \sin \phi \right)\underbrace{(\sqrt{3}\rho \sin \phi \sin \theta)^2}_{3y^2 \ \textrm{in spherical}} \ d\rho \ d\phi \ d\theta$$