# 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:

$$A(y) = \pi (a^2 - y^2)$$

So to find the volume, we need to integrate the function:

$$V = \int_\frac{a}{2}^a \pi (a^2 - y^2) dy$$

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?

Thanks.

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Integrate from $a/2$ to $a$. Note the typos. It should be $x=\sqrt{a^2-y^2}$, so area is $A(y)=\pi(a^2-y^2)$. The integral is mostly right, but $a-h$ should be $a/2$ and there is a missing $dy$. –  André Nicolas Nov 17 '11 at 7:15
Ah sorry. You are right. I will correct it above. Thanks for the help. –  Dylan Nov 17 '11 at 7:16
So not much left to do, integrate, divide by the volume of the sphere, which is the integral from $-a$ to $a$, or twice the integral from $0$ to $a$, or anyway known by you. Note that since we are interested in ratios, we could take $a=1$. –  André Nicolas Nov 17 '11 at 7:29

## 1 Answer

$y=a/2$ 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.

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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I wrote it into the integral as the integration bound after Andre told me where it should go. I initially thought that maybe I needed to replace y with a/2 in the function itself. –  Dylan Nov 17 '11 at 17:27
@Dylan: I see. Then it seems you left the question in an inconsistent state, since the rest of it seems to be working towards something else? Anyway, I'm not sure whether I've answered the question now; or in fact what exactly it was asking after you changed it. –  joriki Nov 17 '11 at 17:32
You and Andre both answered my question. I know how to solve it from here. Thanks. –  Dylan Nov 17 '11 at 17:57