 # A circle x^{2}+y^{2}=a^{2} is rotated around the y-axis to form a solid sphere of radius a. Brenton Dixon 2022-07-23 Answered
Finding volume of a sphere
I am stuck on the following problem:
A circle ${x}^{2}+{y}^{2}={a}^{2}$ is rotated around the y-axis to form a solid sphere of radius a. A plane perpendicular to the y-axis at $y=\frac{a}{2}$ cuts off a spherical cap from the sphere. What fraction of the total volume of the sphere is contained in the cap?
So far I have figured out the following:
Rotating the cap on the y axis we get a height h starting from $y=\frac{a}{2}$. The interval from $y=0$ to $y=\frac{a}{2}$ (the region below the cap) should be:
$a-h$
I also know that the radius of the sliced disk, x, can be derived from the equation of the circle:
$x=\sqrt{{a}^{2}-{y}^{2}}$
Since the area of a circle is $A=\pi {r}^{2}$ the area with respect to y for the circle should be:
$A\left(y\right)=\pi \left({a}^{2}-{y}^{2}\right)$
So to find the volume, we need to integrate the function:
$V={\int }_{\frac{a}{2}}^{a}\pi \left({a}^{2}-{y}^{2}\right)dy$
I know where I should go, but I am not sure what to do about the constraint $y=\frac{a}{2}$ at this point. Should I integrate the terms with respect to y first and then plug in the value which is equal to y? Or should this be done before integrating?
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Step 1
$y=a/2$ is not a constraint; it's one of the integration bounds, and you've already written it into the integral as an integration bound correctly. Now all you have to do is evaluate the integral.
Step 2
By the way, your derivation of the bounds seems unnecessarily complicated. There's no reason to invoke $y=0$, which doesn't play any special role here. The truncated sphere lies between -a and a/2 and the cap that's cut off lies between a/2 and a, so you integrate from a/2 to a; nothing to be subtracted or calculated there.

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