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.

  • $\begingroup$ Why you want to rotate a sphere and not simply a circle? $\endgroup$ – Emilio Novati Feb 25 '16 at 14:55
  • $\begingroup$ @EmilioNovati Why you want to rotate a circle and not simply a sphere? $\endgroup$ – user19405892 Feb 25 '16 at 14:58
  • $\begingroup$ Because it is simpler. $\endgroup$ – Emilio Novati Feb 25 '16 at 15:20
  • $\begingroup$ @EmilioNovati It doesn't really matter, though, in terms of the question but yes we can consider a circle. $\endgroup$ – user19405892 Feb 25 '16 at 15:22

It can be shown that you are describing a circle with radius $1$ and centre $(0,0.5)$.

enter image description here

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 $$\pi \int_{\frac{\sqrt3}2}^{1}(y^2_1-y^2_2)dx$$



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?


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