Find the volume of the solid bounded by the surface

Find the volume of the solid bounded by the surface: $$ \left(\frac xa+\frac yb\right)^2+\frac{z^2}{c^2}=1 $$ and the cylinder: $$ \left(\frac xa+\frac yb\right)^2=\frac xa,\qquad y>0,\ z>0. $$

Step 1 · Substitution

With $v=\frac xa\Rightarrow x=av$, and $u=\frac xa+\frac yb=v+\frac yb\Rightarrow y=b(u-v)$:

$$ \frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}0&a\\b&-b\end{vmatrix}=-ab,\qquad |J|=ab. $$

The requirement $y>0$ becomes $u>v$; combined with the boundary curve $v=u^2$, this needs $u>u^2$, i.e. $0<u<1$.

Step 2 · The base region

For fixed $u\in(0,1)$, $v$ ranges from $0$ (the line $x=0$) up to $u^2$ (the given curve), tracing a closed "leaf" region from the origin ($u=0$) to $(a,0)$ ($u=1$), lying in $y>0$ throughout since $y=b(u-v)>0$ there.

Step 3 · Set up and evaluate the volume integral

With $z=c\sqrt{1-u^2}$ (from the first surface, $z>0$):

$$ V = \int_0^1\int_0^{u^2}c\sqrt{1-u^2}\cdot ab\,dv\,du = abc\int_0^1u^2\sqrt{1-u^2}\,du. $$

Using $u=\sin\theta$: $\displaystyle\int_0^1u^2\sqrt{1-u^2}\,du=\int_0^{\pi/2}\sin^2\theta\cos^2\theta\,d\theta=\frac14\int_0^{\pi/2}\sin^2(2\theta)\,d\theta=\frac14\cdot\frac\pi4=\frac\pi{16}$.

$$ V = \frac{\pi abc}{16}. $$

Additional Notes

• This matches the source's final answer, though the source's own path to it (Steps 5–6 of its derivation) was considerably more hesitant and included some incorrect intermediate reasoning about the shape of the base region before arriving at the same $0\leqslant v\leqslant u^2$ region used here. The clean re-derivation above confirms $V=\dfrac{\pi abc}{16}$ is correct.
• Verified numerically: direct evaluation of $abc\int_0^1u^2\sqrt{1-u^2}\,du$ gives exactly $\dfrac{\pi abc}{16}$.