Solving a surface integral

Question

Given the unit sphere $S = \{(x,y,z)\in \mathbb{R}^3:x^2+y^2+z^2=1\}$. I have to find the value of the surface integral $\iint_S \{ \frac{2x}{\pi} + sin(y^2))\hat{x} + (e^z - \frac{y}{\pi})\hat{y} + (\frac{2z}{\pi} + sin^2y)\hat{z}\}d\sigma$. I used Divergence Theorem to arrive at the answer 3. But the answer is shown to be 4.

Answer 1

You have not shared your working so difficult to say where you made a mistake but $4$ is indeed the correct answer.

$\vec{F} = (\frac{2x}{\pi} + \sin y^2) \hat x + (e^z - \frac{y}{\pi}) \hat y + (\frac{2z}{\pi} + \sin^2 y) \hat z$

$\nabla \cdot \vec F = \frac{2}{\pi} - \frac{1}{\pi} + \frac{2}{\pi} = \frac{3}{\pi}$

$\displaystyle \iint_S \vec F \cdot \hat n \ dS = \iiint \nabla \cdot \vec F \ dV = \frac{3}{\pi} \iiint dV = \frac{3}{\pi} \times \frac{4 \pi}{3} = 4$.

$\iint dV$ is the volume of the unit sphere which we know is simply $\frac{4\pi}{3}$. If you are integrating to find that, you can write the integral in spherical coordiantes as,

$\displaystyle \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^2 \ sin \phi \ d\rho \ d\phi \ d\theta = \frac{4 \pi}{3}$.

Answer 2

Since $\overrightarrow{F} = (\frac{2x}{\pi} + sin y^2, e^z - \frac{y}{\pi}, \frac{2z}{\pi} + sin^2 y)$, $\overrightarrow{\nabla}\cdot\overrightarrow{F} = \frac{2}{\pi} -\frac{1}{\pi} + \frac{2}{\pi} = \frac{3}{\pi}$

Since the volume of the sphere is $\frac{4\pi}{3}$, the answer is 4.