Evaluate $$\iint_{S} (x^3+y^3)\,dy\,dz + (z^3+y^3)\,dz\,dx + (x^3+z^3)\,dx\,dy$$ where $S$ is the surface of the sphere $x^2+y^2+z^2= a^2$.
Spherical coordinates help me in this question, $r^2=a^2$.
We can also write $$\iiint_{S} 2(x^3+y^3+z^3)\,dx\,dy\,dz$$ or not?
Thank you for your advanced ideas.
Based on Mikhail Ostrogradsky a surface integral can be converted to volume integral where P, Q, R are function of x, y, z.
$$\iint_SP\mathop{dy}\mathop{dz} + Q\mathop{dz}\mathop{dx} + R\mathop{dx}\mathop{dy} = \iiint_V(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z})\mathop{dx}\mathop{dy}\mathop{dz}$$
Hence the surface integral becomes
$$\iiint_V(3x^2 + 3y^2 + 3z^2)dx\mathop{dy}\mathop{dz}$$ $$= 3\iiint_V r^2\mathop{dx}\mathop{dy}\mathop{dz}$$ $$= 3\int_{0}^{a}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\pi} r^4cos\phi \mathop{d\theta}\mathop{d\phi}\mathop{dr}$$ $$= 6\pi \int_{0}^{a}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} r^4cos\phi \mathop{d\phi}\mathop{dr}$$ $$= 12\pi \int_{0}^{a} r^4 \mathop{dr}$$ $$= \frac{12}{5}\pi a^5$$