Triple integral $\int_{0}^{2\pi} \int_{0}^{2\cos(\theta)} \int_{0}^{\sqrt{2r\cos(\theta)}} r \ dzdrd\theta$ to find volume of a solid

Question

On evaluating the volume between

$$x^2+y^2 = 2x\\z^2=2x$$

I set up the triple integral $$\int_{0}^{2\pi} \int_{0}^{2\cos(\theta)} \int_{0}^{\sqrt{2r\cos(\theta)}} r \ dzdrd\theta$$

for which Wolfram gives me the answer $\frac{128}{15}$ (which is correct). My work is shown below. I just cannot find the mistake that is producing the wrong result (even though the setup is ok):

$$ \int_{0}^{2\pi} \int_{0}^{2\cos(\theta)} \int_{0}^{\sqrt{2r\cos(\theta)}} r \ dzdrd\theta = \\ \int_{0}^{2\pi} \int_{0}^{2\cos(\theta)} r \sqrt{2r\cos(\theta)} \ drd\theta = \\ \int_{0}^{2\pi} \sqrt{2\cos(\theta)} d\theta \int_{0}^{2\cos(\theta)} r^{3/2} dr = \\ \frac{2}{5} \int_{0}^{2\pi} \sqrt{2 \cos(\theta)} \cdot (2 \cos\theta)^{5/2} d\theta = \\ \frac{2}{5} \int_{0}^{2\pi} cos^3(\theta) d\theta = 0 $$

Answer 1

HINT:

Note that

$$x^2+y^2=2x\implies x\ge 0 \implies \cos(\theta)\ge 0\implies |\theta|\le \pi/$$

SPOLIER ALERT Scroll over the highlighted area to reveal the solution

The integral extends in $z$ from $-\sqrt{2x}$ to $\sqrt{2x}$. So, we can write $$V=\iint_{R_{xy}}2\sqrt{2x}\,dx\,dy$$Now, upon transforming to polar coordinates $(r,\theta)$, we note that the radial variable $r$ extends from $0$ to $2\cos(\theta)$ while the angular variable $\theta$ starts at $-\pi/2$ and ends at $\pi/2$. We can write, therefore $$\begin{align}V&=\iint_{R_{xy}}2\sqrt{2x}\,dx\,dy\\\\&=\int_{-\pi/2}^{\pi/2}\int_0^{2\cos(\theta)}2\sqrt{2r\cos(\theta)}\,r\,dr\,d\theta\\\\&=\int_{-\pi/2}^{\pi/2}\frac45 (2\cos(\theta))^3\,d\theta\\\\&=\frac45 \times 8\times\frac43\\\\&=\frac{128}{15}\end{align}$$as expected!