Volumes of Solids of Revolution

[f1player]

Use the disc method to determine the volume of the solid generated by revolving the region bounded by y = root(x), y=0 and x=4 about the line x=6

I'm completely stuck on this one. I've done it several times now and can't seem to find the right answer. Since there will be a hole in the shape created there will be an inner and outer radius. Am i correct?

I said: inner R = 2
outer R = 6-x = 6-y^2

is this where i'm wrong?

 

[soroban]

Hello, f1player!

Use the disc method to find the volume of the solid generated by revolving
the region bounded by: \(\displaystyle y\,=\,\sqrt{x},\:y\,=\,0,\:x\,=\,4\) about the line \(\displaystyle x\,=\,6\)

I said: inner \(\displaystyle R\:=\:2\)
. . . . .outer \(\displaystyle R\:=\:6\,-\,x\:=\:6\,-\,y^2\;\) . . . This is correct

There are still a number of possible errors in the set-up.

The integral should be: \(\displaystyle \:\pi\L\int\)\(\displaystyle \left[(6-y^2)^2\,-\,2^2\right]\,dy\)

The limits should be: \(\displaystyle \,y\,=\,0\) to \(\displaystyle y\,=\,2\)

The volume will be: \(\displaystyle \:\left[\pi\left(32y\,-\,4y^3\,+\,\frac{y^5}{5}\right)\,\right]^{^2}_{_0}\;=\;\L\frac{192\pi}{5}\)

 

[f1player]

Yes, thanks for that

Turns out I got the limits wrong

 

[galactus]

Just for kicks, we can do this with shells also.

\(\displaystyle \L\\2{\pi}\int_{0}^{4}(6-x)\sqrt{x}dx=\frac{192{\pi}}{5}\)

Actually, that's easier than discs. Probably why they wanted you to use discs.