Volume of solid bounded by $z^2 = x^2 + y^2$ and $x^2 + y^2 = 2x$

Question

Calculate the volume of the solid bounded by $z^2 = x^2 + y^2 $ and $x^2 + y^2 = 2x$

My attempt:

Using cylindrical coordinates,

$$ \mathrm{Vol} = \int_{-\pi/2}^{\pi/2} \int_0^1 \int_{-r}^{r} r \,dz\, dr\, d\theta $$

But the answer is $\frac{64}{9}$

I would be glad if someone could point my mistake and give a solution.

Thanks in advance.

Answer 1

$x^2+y^2=2x\implies (x-1)^2+y^2=1$ So it is a circle on $x,y$ plane centered the $(1,0)$.

We can find the volume by double integral on the region $D$ enclosed by the circle.

Assume $z\geq0$ and by symetry we should multiply by $2$.

Thus, $$V=2\int\int_D \sqrt{(x^2+y^2)} \, dx \, dy$$

To make it easier we may use polor cordinate but be careful our cirle has center at $(1,0)$.

$$ V=2 \int_{-\pi/2}^{\pi/2}\int_0^{2cos(\theta)} \sqrt{r^2} r \, dr \, d \theta $$

We are done.

Note: In your solution $r$ should change from $0$ to $2cos(\theta)$