Solid enclosed by the paraboloid $\frac{y^2}{b^2}+\frac{z^2}{c^2}=2\frac{x}a$ and the plane $x=a.$ 
Compute the volume of the solid enclosed by the paraboloid $$\frac{y^2}{b^2}+\frac{z^2}{c^2}=2\frac{x}a$$ and the plane $x=a.$

My attempt:
I considered slices of the solid parallel with the plane $x=a$ and summed up their areas.
$$\begin{aligned}y&=br\sqrt{\frac2a}\sin\varphi\\z&=cr\sqrt{\frac2a}\cos\varphi,\\\varphi&\in[0,2\pi],\\r&\in[0,\sqrt x],\\x&\in[0,a]\\\text{Jacobian 
: } J_\psi&={2bc r}a\\\int_0^a\int_0^{2\pi}\int_0^{\sqrt x}\frac{2bcr}adrd\varphi dx&=\frac{2bc}a2\pi\int_0^a\frac{x}2dx\\&=\frac{2bc}a2\pi\frac12\int_0^axdx\\&=\frac{2bc}a\pi\frac{a^2}2\\&=abc\pi\end{aligned}$$
Is this right and if so, is there any other efficient method?
 A: If you take a section at $x = t$, then the area of the section is
$A(t) =\pi b c \left(\dfrac{2t}{a} \right)$
Now the volume is
$ V = \displaystyle \int_{t = 0}^{t = a} A(t) dt = a b c \pi $
A: Yes, your approach is correct. Here we can see another approach slightly different.

Setting the solid:
$$E=\left\{(x,y,z): (y,z)\in D, \frac{a}{2}\left( \frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\right)\leqslant x\leqslant a \right\}, \quad D=\left\{(y,z): \frac{y^{2}}{(\sqrt{2}b)^{2}}+\frac{z^{2}}{(\sqrt{2}c)^{2}}=1\right\}.$$
Using the change of variables,
$$y=\sqrt{2}b\cos\theta$$
$$z=\sqrt{2}c\sin\theta$$
with $\theta\in [0,2\pi[$ and $r\in [0,1]$. Then the Jacobian transformation is given with determinant $2bcr$ and we get also the volume of solid $E$ as follows:
\begin{align*}
V(E)&=\iiint_{E}\,{\rm d}V,\\
&=\iint_{D}\left[\int_{\frac{a}{2}\left(\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} \right)}^{a}\, {\rm d}x \right]\, {\rm d}A,\\
&=\iint_{D}\left[a-\frac{a}{2}\left(\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\right)\right]\, {\rm d}A,\\
&=\int_{0}^{2\pi}\int_{0}^{1}-\left[\frac{a}{2}\left(\frac{(\sqrt{2}br\cos\theta)^{2}}{b^{2}} +\frac{(\sqrt{2}cr\sin \theta)^{2}}{ c^{2}}\right)-a\right]\cdot 2bcr\, {\rm d}r{\rm d}\theta,\\
&=\int_{0}^{2\pi}\int_{0}^{1}a(1-r^{2})\cdot 2bcr\, {\rm d}r{\rm d}\theta,\\
&=\int_{0}^{2\pi}\frac{1}{2}abc\, {\rm d}\theta,\\
&=+abc\pi
\end{align*}
Also, notice that if $$E=\left\{(x,y,z): (y,z)\in D, a\leqslant x\leqslant \frac{a}{2}\left( \frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\right)\right\},$$we get $V(E)=-abc\pi$ so this also gives us conditions on the constants $a,b,c$ for the problem to make sense in the physics.
