Question

Volumes and rotating!

Answer 1

Let R be the region bounded by y=sin(x) on the interval [0,pi]
a) What is the volume of the solid when rotated about the x-axis?
b) What is the volume of the solid when rotated about the y-axis?

Answer 2

\[Area=\pi r^2\] a)\[\pi \int\limits_{0}^{\pi}(\sin(x))^2dx\] b)\[\pi \int\limits_{-1}^{1} (\arcsin(x))^2dx\]
These are the equations I came up with..are they correct?

Answer 3

The first one is definitely right, and the second one is not quite right though, since you'll have to break it up into two parts, plus arcsine is a hassle to integrate. The approach isn't wrong, it's just there's another alternate and easier way to set it up.

The way you're doing is called the "disk" or "washer" method because geometrically each infinitesimal volume layer you add on with your integral looks like a bunch of disks stacked up, but there's another method called the shell method. The shell method constructs your volume with a bunch of concentric shells, like a bunch of tin cans inside of each other. A picture might help haha


So if you look at this surface area of the cylinder taken from a piece of our sine wave, it has this area:

\[A=2 \pi x * \sin x\]
So if you give this area a little width it becomes a little volume:

\[dV = Adx = 2\pi x \sin x dx\]
and you can now add up all the infinitely small pieces up from 0 to \(\pi\) to get the volume.

Errr hopefully that's not too long winded and makes some sense? :O