Exercise: Calculate the volume between $ x\geq 0, y \geq 0, z\geq 0, y^2 + z^2 =1, x=2y$.
I know it's a routine exercise but I fail to draw a proper graph, so my result may be wrong. Can someone check it?
Attempt: $$ V = \iint_{pr(yOz)} \int_{0}^{2y} dxdydz=\iint_{pr(y0z)}2y\ dydz = \ldots = \frac{2}{3} $$
The double integral was calculated with polar coordinates for $ r\in [0,1] $ and $\theta \in [0,\frac{\pi}{2}] $.
First, it is easier for the polar coordinates conversion to switch $x$ and $z$. Then you have $x^2+y^2=1$ and $z=2y$. Then the triple integral becomes $$\int_0^{\pi/2} \int_0^1 \int_0^{2\sin\theta} r dz dr d\theta$$