evaluating $\iiint\limits_E x \; dV$; $E$ between $x=ay^2+az^2$ and $x=a$

Question

$\iiint\limits_{x=ay^2+az^2}^{x=a} \!\!\!\!\! x \; dV$ (is writing integral limits like this fine?)
in my head this volume looks like an x-axis paraboloid up to the unit circle
I try to use cylindrical coordinates along the x axis with $x,\; \theta, \text{ and } r$
my limits then look like: $\int\limits_{0}^{2\pi}\int\limits_0^a\int\limits_0^{\sqrt{\frac{x}{a}}}x\;dr\;dx\;d\theta$
then evaluating one integral at a time I get $\int\limits_{0}^{2\pi}\int\limits_0^a\int\limits_0^{\sqrt{\frac{x}{a}}}x\;dr\;dx\;d\theta= \int\limits_{0}^{2\pi}\int\limits_0^a \frac{1}{\sqrt{a}}x^{\frac{3}{2}} \;dx\;d\theta= \int\limits_{0}^{2\pi} \frac{2}{5\sqrt{a}}a^{\frac{5}{2}} \;d\theta= \int\limits_{0}^{2\pi} \frac{2}{5}a^2 \;d\theta= \frac{4\pi a^2}{5} $

plugging it into a online calculator using rectangular coordinates gives $\frac{\pi a^2}{3}$, so I deff did something wrong somewhere in there

Answer 1

If you take slices of the paraboloid at $x$ where $0 \le x \le a $, then the cross section you will obtain is

$ x = a (y^2 + z^2) $

which is a circle whose squared radius is $\dfrac{x}{a} $, and therefore its area is $ \dfrac{\pi x}{a} $, and hence $dV = \dfrac{\pi x dx}{a}$

The integral then is

$ I = \displaystyle \int_{0}^{a} \dfrac{\pi x^2 dx}{a} = \dfrac{ \pi a^2}{3} $

Answer 2

Here is a plot of such region using Mathematica when $a = 1$:

Hopefully this helps!