Find the volume of the solid generated by rotating $B$ about $l$ 
In a space, let $B$ be a sphere (including the inside) with radius of $1$. Line $l$ intersects with $B$, the length of the common part is the line segment with the length of $\sqrt{3}$. Find the volume of the solid generated by rotating $B$ about $l$.

This solid looks like a torus pushed together. Thus, if we take the volume of the cylinder formed which here is $8\pi$ and subtract the inner part, which is a sector of a sphere rotated about the axis of the $\sqrt{3}$ line we get the desired volume. But I struggle how to find such a volume. Maybe there is a different way to find the volume I am not seeing that is easier.
 A: It can be shown that you are describing a circle with radius $1$ and centre $(0,0.5)$.

There are two parts to the curve:
Green: $y_1=\frac 12 + \sqrt{1-x^2}$
Red: $y_2=\frac 12 - \sqrt{1-x^2}$
Rotate the green part about the $x$-axis between $x=-1$ and $x=1$ to find a large volume.
Then rotate the red part between $x=-1$ and $x=-\frac{\sqrt3}2$ and between $x=\frac{\sqrt3}2$ and $x=1$ to get two smaller parts. Subtract the smaller bits from the larger. 
$y_1^2=\frac 14 + \sqrt{1-x^2}+1-x^2$
$y_2^2=\frac 14 - \sqrt{1-x^2}+1-x^2$
Volume between $x=0$ and $x=\frac{\sqrt3}2$ is $$\pi \int_0^{\frac{\sqrt3}2}y^2_1 dx$$
Volume between $x=\frac{\sqrt3}2$ and $x=1$ is $$\pi \int_{\frac{\sqrt3}2}^{1}(y^2_1-y^2_2)dx$$
A: Hint:
The resulting solid of revolution is the same as rotating a circle of radius $1$ around an axis that intersect on the circle a segment of lenght $\sqrt{3}$.
If the circle has equation $x^2+y^2=1$ the axis can be the straight line $x= \frac{1}{2}$.  So the volume is given by circular shells of radius $\frac{1}{2}$, thickness $dx$ and height $2\sqrt{1-x^2}$ for $-1\le x\le \frac{1}{2}$ and is the integral:
$$
4\pi\int_{-1}^{\frac{1}{2}}(\frac{1}{2}-x)\sqrt{1-x^2}dx
$$
Can you do from this?
