Bounded by the cylinder $y^{2}+ z^{2}=4$ and the planes x = 2y, x=0, z=0 in the first octant. I know how to solve this in the cartesian coordinate.

Question

Bounded by the cylinder $$y^{2}+ z^{2}=4$$ and the planes x = 2y, x=0, z=0 in the first octant. I know how to solve this in the cartesian coordinate. and the correct answer is $\frac{16}{3}$

but when I attempt to use the polar coordinate, I didn't get the answer correct.

$$\int_{0}^{\pi/2}\int_{0}^{2}\int_{0}^{4sin(\theta)}r \ dx \ dr \ d\theta$$ and I got the answer 8.

Would you let me know what went wrong?

Answer 1

Variety of $x$ depends on both $r$ and $\theta$, so with $x=2y=2r\sin\theta$ we should have $$\int_{0}^{\pi/2}\int_{0}^{2}\int_{0}^{2r\sin\theta}r \ dx \ dr \ d\theta=\frac{16}{3}$$