The equation for the ellipse states $x^2+y^2-2x=0$ (...) from which I have $\rho(\theta)= \displaystyle\frac{\cos\theta}{1+\cos^2\theta}$ with $0\leq\theta\leq\pi$.
The equation of the ellipse is
\begin{equation*}
2x^{2}+y^{2}-2x=0
\end{equation*}
from which I've obtained
\begin{equation*}
\rho (\theta)=\frac{2\cos \theta }{\cos ^{2}\theta +1},\qquad \text{with }-\pi/2 \leq
\theta <\pi/2,
\end{equation*}
because the tangent to the ellipse at $(x,y)=(0,0)$ is the vertical line $x=0$. With these corrections the volume integral becomes
\begin{eqnarray*}
V &=&\int_{-\pi /2}^{\pi /2}\left( \int_{0}^{\frac{2\cos \theta }{\cos
^{2}\theta +1}}2\rho \sqrt{1-\rho ^{2}}d\rho \right) \,d\theta \\
&=&-\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left. \left( 1-\rho ^{2}\right)
^{3/2}\right\vert _{0}^{\frac{2\cos \theta }{\cos ^{2}\theta +1}}\,d\theta
\\
&=&-\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( 1-\left( \frac{2\cos \theta }{
\cos ^{2}\theta +1}\right) ^{2}\right) ^{3/2}-1\,\ d\theta
\end{eqnarray*}
Simplifying, we have that
\begin{eqnarray*}
V &=&\frac{2}{3}\pi -\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( \left( \frac{
\sin ^{2}\theta }{\cos ^{2}\theta +1}\right) ^{2}\right) ^{3/2}\,d\theta \\
&=&\frac{2}{3}\pi -\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( \frac{\sin
^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta \\
&=&\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin ^{2}\theta
}{\cos ^{2}\theta +1}\right) ^{3}\,d\theta ,
\end{eqnarray*}
because the integrand is an even function. The remaining integral can be
converted into a rational function of $t=\tan \frac{\theta }{2}$ by the
Weirstrass substitution $t=\tan \frac{
\theta }{2}$, which I then evaluated in SWP (Scientific Work
Place):
\begin{eqnarray*}
V &=&\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin
^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta ,\qquad t=\tan \frac{
\theta }{2} \\
&=&\frac{2}{3}\pi -\frac{64}{3}\int_{0}^{1}\frac{t^{6}}{\left(
1+t^{4}\right) ^{3}\left( 1+t^{2}\right) }\,dt \\
&=&\cdots \\
&=&\frac{4}{3}\pi -\frac{7}{12}\sqrt{2}\pi =\frac{16-7\sqrt{2}}{12}\pi .
\end{eqnarray*}
ADDED. The integrand in $t$ can be expanded into partial fractions as follows
\begin{equation*}
\frac{t^{6}}{\left( 1+t^{4}\right) ^{3}\left( 1+t^{2}\right) }=\frac{1}{8}
\frac{t^{2}-1}{1+t^{4}}+\frac{1}{4}\frac{3+t^{2}}{\left( 1+t^{4}\right) ^{2}}
-\frac{1}{2}\frac{1+t^{2}}{\left( 1+t^{4}\right) ^{3}}-\frac{1}{8\left(
1+t^{2}\right) }.
\end{equation*}
ADDED 2. WolframAlpha computation of $V =\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta$ confirms the result above.
The equation of the sphere is $r^2+(z-2)^2=4$, so $z=2 \pm \sqrt{4-r^2}$. The solid of interest is bounded below and above by the sphere and by the cylinder $r=1$ on the side.
Then,
$$V=\int_{0}^{2\pi} \int_{0}^{1} \int_{2-\sqrt{4-r^2}}^{2+\sqrt{4-r^2}} r dz dr d\theta$$
$$=\int_{0}^{2\pi} \int_{0}^{1} 2r\sqrt{4-r^2} dr d \theta$$
$$=-2\pi \int_{4}^{3} u^{\frac{1}{2}} du$$
$$=\frac{4}{3}\pi (8-3\sqrt{3})$$
Best Answer
Let $I = \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta$
$I = \frac{-1}{2}\int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ d(4-r^2) \ d \theta$
$I = \frac{-1}{2}.\frac{2}{3}\int_0^{\pi/2}\big[(4-r^2)^{3/2}\big]^{2cos\theta}_0d\theta$
$I = \frac{-1}{2}.\frac{2}{3}\int_0^{\pi/2}\big[(4-4cos^2\theta)^{3/2} - 4^{3/2}\big]d\theta$
$I = \frac{-1}{3}.2^3\int_0^{\pi/2}\big[(sin^2\theta)^{3/2} - 1\big]d\theta$
$I = \frac{-1}{3}.2^3\int_0^{\pi/2}\big[(sin^3\theta)- 1\big]d\theta$
$\bigg[ \int^{\pi/2}_0sin^mxdx = \frac{m-1}{m}.\frac{m-3}{m-2}...1 \bigg]$, if m is odd
$I = \frac{-2^3}{3}\bigg[\big[\frac{2}{3}.1\big] -\frac{\pi}{2}\bigg]$
Thus, $I = \frac{8}{18}[3\pi - 4] = \frac{4}{9}[3\pi - 4]$
Can you proceed now?