Q: Find the volume cut off from the cylinder $x^2+y^2=ax$ by the planes $z=0 $ and $z=x$
Given Answer:$\frac{128a^3}{15}$
My answer:$\frac{\pi a^3}{4}$
Working:
So, first off, we can decipher from the question that $z$ will vary from $0$ to $x$.
$y$ will vary from $-\sqrt{ax-x^2}$ to $\sqrt{ax-x^2}$.
$x$ will vary from $0$ to $a$
So, the integral to be evaluated will be: $$\int_{0}^a\int_{-\sqrt{ax-x^2}}^{\sqrt{ax-x^2}}\int_{0}^x 1\cdot dzdydx$$ $$=\int_{0}^a2x\sqrt{ax-x^2}dx$$
$$=a\int_{0}^a\sqrt{ax-x^2}dx - \int_{0}^a(a-2x)\sqrt{ax-x^2}dx$$
The second integral turns out to be $0$, so we're left with only the first integral $$=a\int_{0}^a\sqrt{ax-x^2}dx$$ $$=a\int_{0}^a\sqrt{\left(\frac{a}{2}\right)^2-\left(x-\frac{a}{2}\right)^2}dx$$ $$=a\left[\frac{x-\frac{a}{2}}{2}\sqrt{ax-x^2}+\frac{a^2}{4}\sin^{-1}\left(\frac{x-\frac{a}{2}}{\frac{a}{2}}\right)\right]_0^a=\frac{\pi a^3}{4}$$
The calculation looks to be fine, so I think I might be messing up with the limit calculation. If someone could verify, it'd be of great help. Thank you!!
$ \displaystyle \frac{128a^3}{15}$ is incorrect answer. It should be $\displaystyle \frac{\pi a^3}{8}$. There is a small mistake in your working in the last step -
$V = \displaystyle a\left[\frac{x-\frac{a}{2}}{2}\sqrt{ax-x^2}+\frac{a^2}{\color {blue} 4}\sin^{-1}\left(\frac{x-\frac{a}{2}}{\frac{a}{2}}\right)\right]_0^a$
$4$ is incorrect. It should be $8$. See below -
$V = \displaystyle a\left[\frac{x-\frac{a}{2}}{2}\sqrt{ax-x^2}+\frac{(a/2)^2}{2}\sin^{-1}\left(\frac{x-\frac{a}{2}}{\frac{a}{2}}\right)\right]_0^a = \frac{\pi a^3}{8}$
Using cylindrical coordinates,
$x = r \cos\theta, y = r \sin\theta, z = z$
$x^2 + y^2 \leq ax \implies r \leq a \cos\theta$
Also note for the parametrization we are using, $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$
So integral becomes
$V = \displaystyle \int_{-\pi/2}^{\pi/2} \int_0^{a\cos\theta} \int_0^{r\cos\theta} r \ dz \ dr \ d\theta = \frac{\pi a^3}{8}$