Volume of a sphere cut out by a cylinder

Question

The goal is to find a formula for the volume of a sphere $ x^2 + y^2 + z^2 \le R^2 $ cut by a cylinder $ (x - \frac{R}{2})^2 + y^2 \le \frac{R^2}{4} $
I was able to solve it using double integrals as follows:
The area of integration $D$ is
$ -\frac{\pi}{2} < \varphi < \frac{\pi}{2} $
$ 0 < r < R\cos\varphi $
and I am using polar coordinates to describe it
$ \left\{\begin{matrix} x=r\cos\varphi \\ y=r\sin\varphi \end{matrix}\right. $
I solve for $z$:
$ z=\pm \sqrt{R^2-r^2} $
I get the following integral
$$ V=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\varphi\int_{0}^{R\cos\varphi}\sqrt{R^2-r^2}rdr=$$ $$=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[-\frac{1}{3}(R^2-r^2)^{\frac{3}{2}}]_{0}^{R\cos\varphi}d\varphi=$$ $$=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\frac{1}{3}R^3-\frac{1}{3}R^3\sin^3{\varphi})d\varphi=$$ $$=\frac{2}{3}R^3[\varphi]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} - \frac{2}{3}R^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^3{\varphi}d\varphi=$$ $$=\frac{2}{3}\pi R^3-\frac{2}{3}R^3[\frac{\cos^3{\varphi}}{3}-\cos(\varphi)]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=$$ $$=\frac{2}{3}\pi R^3 $$
And everything would be fine, if not for the fact that the formula for volume of a sphere is $\frac{4}{3}\pi R^3$. On the below image I am clearly seeing room to fit a second symmetric cylinder on the opposite side of $yOz$ plane (grey one):
And my formula already adds up to $\frac{4}{3}\pi R^3$, with plenty of sphere's volume to spare. Where is the mistake?

Answer 1

(1) Issue 1 : You are actually calculating half the Volume of the Sphere , to get what you got , hence Doubling it will get the Volume of the Sphere.
(2) Issue 2 : You are not calculating the Intersection with Cylinder.

(A) You are using $\varphi$ to get $r$ in terms of $R$ , along the $Z$ Direction , which is right.
More-over , you have to take that between $\pm pi/2$ , which is also right.

(B) You are then using the Same $\varphi$ to calculate $x$ & $y$ which is not right.
You have to use $\theta$ , along the $XY$ Plane , which will be between $\pm \pi$.
If you had used that , or used the limits $\pm \pi$ , you would have got the Volume of the whole Sphere.
Having used half the limits , you got half the Volume of the Sphere.

(C) Okay , then use $x = r \cos \theta$ & $y = r \sin \theta$ , to resolve Issue 1.
IMPORTANT : That is not Polar Co-ordinates. We have to use Spherical Co-Ordinates.

(D) You have to then handle the Issue 2 where you make the limits between the Cylinder.
You will get the tine slice along the $XY$ Plane , which will vary with the $Z$ Direction.

I will leave it till here. You might want to try to complete the Integral. In Case you have Difficulties , I will try to update this Post by tomorrow.