# Find the volume of a solid which is bounded by the paraboloid $4z=x^2 + y^2$, the cone $z^2 = x^2 + y^2$ and the cylinder $x^2 + y^2 = 2x$

I approached this problem by trying to find the volume bounded by the paraboloid and the cylinder and then subtracting it from the volume bounded by the cone and the cylinder. But I am getting the wrong answer. I converted the all the bounds into cylindrical co-ordinates.

For finding the volume bounded by the cone and the cylinder,

Bounds of integration: $$\ z=0 \$$ to $$\ z=r \ \ , \ \ r=0 \$$ to $$\ r=2 \cos(\theta) \ \ , \ \ \theta=0 \$$ to $$\ \theta=2 \pi \$$

For finding the volume bounded by the paraboloid and the cylinder,

Bounds of integration: $$\ z=0 \$$ to $$\ z= \frac{r^2}{4} \ \ , \ \ r=0 \$$ to $$\ r=1 \ \ , \ \ \theta=0 \$$ to $$\ \theta=2 \pi \$$

Please check if the bounds are correct. Thanks.

Your bounds for $$\theta$$ are wrong; $$\theta$$ should run from $$-\frac{\pi}{2}$$ to $$\frac{\pi}{2}$$: $$V=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\cos\theta}\int_{\frac{r^2}{4}}^rr\,dz\,dr\,d\theta=\frac{32}{9}-\frac{3 }{8}\pi.$$ If you want to do it by subtracting the volume bouded by the paraboloid to the one bouded by the cone, then \begin{align*} V&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\cos\theta}\int_{0}^rr\,dz\,dr\,d\theta-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\cos\theta}\int_{0}^{\frac{r^2}{4}}r\,dz\,dr\,d\theta=\\ &=\frac{32}{9}-\frac{3}{8}\pi. \end{align*}