# Triple integral over a sphere with cylindrical coordinates

I have the sphere $C: x^2+y^2+z^2=1$ and I have to calculate $\int \int \int 2z+x \, dz \, dy \, dx$

In cylindrical coordinates I have $$x=r\cos \theta$$ $$y=r\sin \theta$$ $$z=z$$

$$0\leq r\leq 1$$ $$0\leq \theta\leq 2\pi$$ $$-1\leq z\leq 1$$ $$dxdy=rdrd\theta$$

Then my integral is becoming $$\int_{-1}^1\int_0^{2\pi}\int_0^1(2z+r\cos\theta)r \, dr \, d\theta \, dz$$

Is the setup of the integral with cylindrical coordinates correct or there is a mistake in the process?

No, because that setup causes you to integrate over the cylinder $r<1,-1<z<1$. You need to incorporate the condition that $$x^2+y^2<1-z^2,$$ (which says you are inside the sphere; the way you do this is to note that $x^2+y^2=r^2$, so the condition can be written $$r<\sqrt{1-z^2},$$ which you incorporate into the limits of the $r$-integral, so the inner integral operator is $$\int_0^{\sqrt{1-r^2}} \big(\cdots\big) \, r \, dr.$$ Everything else, the outer two integrals and the substitution into the integrand and volume element, looks fine.