I tried solving this problem:
Evaluate $\iint_Sx^3dydz+x^2ydzdx+x^2zdxdy$
Where S is the surface bounded by $z=0, z=b, x^2+y^2=a^2$
Using Green Theorem, $\iint_Sx^3dydz+x^2ydzdx+x^2zdxdy=\iiint_V\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}=\iiint_V5x^2dxdydz$
Inserting the intervals gives: $4\int_0^a\int_0^{\sqrt{a^2-x^2}}\int_0^b5x^2dxdydz$
I multiplied the integral by 4 since there are four quadrants.
The problem is now to calculate $4\int_0^a\int_0^{\sqrt{a^2-x^2}}\int_0^b5x^2dxdydz$
I got $10a^4b\pi$
However the textbook I'm studying gave a contrary answer: $\frac{5}{4}\pi a^4b$
It is easier to evaluate the integral in cylindrical coordinates. We have
$$\int_0^b \int_0^{2\pi} \int_0^a 5\rho^2\cos^2\phi \rho \,d\rho\, d\phi\, dz=5\pi b\frac{a^4}{4} \tag 1$$
where we used $\cos^2\phi =\frac{1+\cos 2\phi}{2}$ and $\int_0^{2\pi}\cos 2\phi d\phi=0$ to evaluate the integral in $(1)$
In Cartesian coordinates, we can evaluate the volume integral as
$$\begin{align} 4\int_0^b \int_0^a\int_{0}^{\sqrt{a^2-y^2}}5x^2dx\,dy\,dz&=20b\int_0^a\,\frac13 (a^2-y^2)^{3/2}\\\\ &=20b\frac13 \left(\frac{3\pi a^4}{16}\right)\\\\ &=\frac54 \pi a^4b \end{align}$$
as expected!