$\int_{S}^{} z^{2} dS$ , $S$ sphere

Question

$S$ is a sphere of radius $R$. Using, if you want to, some symmetrical argument, find $\int_{S}^{} z^{2} dS$.

I used spherical coordinates: $$x=\rho \sin(\varphi )\cos(\vartheta )$$ $$y=\rho \sin(\varphi )\sin(\vartheta )$$ $$z= \rho \cos(\varphi)$$ with restrictions $$0\leq \rho \leq R$$ $$0\leq \varphi \leq\pi$$ $$0\leq \vartheta \leq2\pi$$ so I took the triple integral $$\int_{0}^{R}\int_{0}^{\pi }\int_{0}^{2\pi} \rho ^{2} (\cos(\varphi))^{2}\rho ^{2} \sin(\varphi)d\vartheta d\varphi d\rho$$

and I got as an answer the value $$\frac{4\pi}{15} R^{5}$$.

I am not sure if this answer is correct. I have a sphere of radius $R$ but I took specifically the sphere $ x^{2} + y^{2} + z^{2} = R^{2} $ Is this a fault?

Answer 1

Thanks to @Vercassivelaunos I understood the exercise. I thought that I have to use the whole ball. So $$x=R \sin(\varphi )\cos(\vartheta )$$ $$y=R \sin(\varphi )\sin(\vartheta )$$ $$z=R \cos(\varphi)$$ with restrictions $$0\leq \varphi \leq\pi$$ $$0\leq \vartheta \leq2\pi$$

and I have the surface $$\Phi (\varphi, \vartheta)= ( R \sin(\varphi )\cos(\vartheta ), R \sin(\varphi )\sin(\vartheta ) , R \cos(\varphi) ) $$

$$\left \| \Phi _{\varphi}\times \Phi _{\vartheta} \right \| = R^{2} \sin(\varphi)$$ so I get $$\iint_{S}^{} z^{2} dS = \int_{0}^{\pi} \int_{0}^{2\pi} R^{2}(\cos(\varphi))^{2} R^{2} \sin(\varphi) d\vartheta d\varphi $$ which gives me the value $$\frac{4}{3} \pi R^{4}$$

Answer 2

$$ {1 \over 3}\, R^2\ \underbrace{\int R^2\,\mathrm{d}\Omega}_{4\pi R^2} $$