# Math Help - Integration #2

1. ## Integration #2

a) Find the area of the circle formed when a sphere is cut by a plane at a distance y from the centre, where y > r.
I worked this out to be $\pi(r^2 - y^2)$.
b) By integration, prove that the volume of a 'cap' of height $\frac{1}{4}r$ from the top of the sphere is $\frac{{11\pi}r^3}{192}}$.
I'm not really sure what to do here. I can find the area of the circular base of the 'cap', but I don't know what to do with it.

Thanks.

2. ## Re: Integration #2

What you might want to do is take the area of a vertical cross-section of the cap and rotate that through $2 \pi$ radians -- google up on "solid of revolution" for background. When I was at school (hrumphity years ago) it was taught in advanced applied mathematics at the grade one is at when one is aged 17-18.

3. ## Re: Integration #2

That would work, but how could I find the value of the base of that cross-section?

4. ## Re: Integration #2

Originally Posted by Fratricide
a) Find the area of the circle formed when a sphere is cut by a plane at a distance y from the centre, where y > r.
I worked this out to be $\pi(r^2 - y^2)$.
I assume you mean that r is the radius of sphere and y< r, not y> r.

b) By integration, prove that the volume of a 'cap' of height $\frac{1}{4}r$ from the top of the sphere is $\frac{{11\pi}r^3}{192}}$.
I'm not really sure what to do here. I can find the area of the circular base of the 'cap', but I don't know what to do with it.

Thanks.
Think of the "cap" as a series of disks at distance "y" from the center of the sphere, y going from (1/4)r to r, with thickness "dy". What is the area of a circle at distance y from the center of the sphere? What is the volume of such a disk?

5. ## Re: Integration #2

Is the area simply what I had in part (a)? And the volume the area multiplied by the thickness, dy? Are you hinting at something like $\pi{\int_{{\frac{1}{4}r}}^{r}(r^2 - y^2)^2 dy}$ perhaps?