Finding $\int_{S}^{} x^{4} \sin (x^{3}z^{5})\,dx\,dy\,dz$ where $S$ is part of a sphere

Question

Let $S$ be the subset of the sphere $x^{2} + y^{2} + z^{2} = 1, z > 0$. Calculate the integral $$\int_{S}^{} x^{4} \sin (x^{3}z^{5})\,dx\,dy\,dz$$

So I know that this is a surface integral. I used these parameters:

$$\boldsymbol{\mathbf{}\Phi} (\varphi ,\theta )=(\sin \varphi \cos \theta, \sin \varphi \sin \theta, \cos \varphi) , 0<\varphi < \frac{\pi}{2}, 0<\theta<2\pi$$ I also found $$\left \| \Phi_{\phi} \times \Phi_{\theta} \right \| = \sin \varphi$$ So I got the double integral $$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi} \sin^{4}\varphi\cos^{4}\theta \sin(\sin^{3}\varphi \cos^{3}\theta \cos^{5}\varphi)\sin\varphi \,d\varphi \,d\theta $$ but I don't think that it's a good idea.

Answer 1

Denote $f(x,y,z) = x^{4} \sin (x^{3}z^{5})$. We have for all $(x,y,z) \in S$

$$f(x,y,z) = -f(-x,y,z)$$ and therefore

$$\int_{S}^{} x^{4} \sin (x^{3}z^{5})dxdydz=0.$$

Answer 2

Note

$$\int_{S}^{} x^{4} \sin (x^{3}z^{5})dxdydz \\= \int_{S, x>0}^{} x^{4} \sin (x^{3}z^{5})dxdydz+ \int_{S, x<0}^{} x^{4} \sin (x^{3}z^{5})dxdydz\\ = \int_{S, x>0}^{} x^{4} \sin (x^{3}z^{5})dxdydz- \int_{S, x>0}^{} x^{4} \sin (x^{3}z^{5})dxdydz=0 $$