# Volumes/Integration (I think it's called Volumes of Revolution) I get a negative sign

• February 11th 2010, 05:57 PM
s3a
Volumes/Integration (I think it's called Volumes of Revolution) I get a negative sign
I'm referring to #9 of the "mywork.pdf" file. I get the correct magnitude of the final answer but I get the wrong sign (a negative). I know that volume cannot be a negative value so I was hoping someone could tell me why I am wrong. If I just ruined the order of things, simply notyfing me of that would be ok since I think I'm able to solve the algebra by myself...unless my issue is an algebra mistake in which case I'd need my mistake explained.

Any help would be greatly appreciated!
• February 11th 2010, 06:10 PM
skeeter
$V = \pi \int_1^2 \left(\frac{2}{x}\right)^2 \, dx$
• February 11th 2010, 06:27 PM
s3a
Did you look at #9? If so; don't you mean [0,1] instead of [1,2] ? And where did 2/x come from? Sorry I know I said I'd be able to figure it out but I didn't think it would look this different.
• February 11th 2010, 06:37 PM
Quote:

Originally Posted by s3a
I'm referring to #9 of the "mywork.pdf" file. I get the correct magnitude of the final answer but I get the wrong sign (a negative). I know that volume cannot be a negative value so I was hoping someone could tell me why I am wrong. If I just ruined the order of things, simply notyfing me of that would be ok since I think I'm able to solve the algebra by myself...unless my issue is an algebra mistake in which case I'd need my mistake explained.

Any help would be greatly appreciated!

the reason you get a negative value is because the volume of revolution
of $\sqrt{x}$ is a smaller volume than the volume of revolution of $x$ from x=0 to x=1.

You've subtracted the volumes the wrong way around!

i assume you are calculating the volume of revolution of the bounded region between x=0 and x=1.

It's true that to calculate the area between the curves, we subtract the lower function from the upper one.
However, as these are being revolved the lower one (as they are both below y=1) will trace out a larger volume.
There is an error in your integration but is inconsequential since powers of 1 are 1.
• February 11th 2010, 06:44 PM
skeeter
Quote:

Originally Posted by s3a
Did you look at #9? If so; don't you mean [0,1] instead of [1,2] ? And where did 2/x come from? Sorry I know I said I'd be able to figure it out but I didn't think it would look this different.

sorry ... I just instinctively looked at the first one.
• February 11th 2010, 07:00 PM
drumist
When doing your calculation, you should be doing something like $V=\pi \int_0^1 \left[ R^2(x) - r^2(x) \right] dx$, where R(x) is the radius of the larger circle (outside circle) and r(x) is the radius of the smaller circle. Notice when rotating about the line y=1, the curve $y=\sqrt{x}$ is actually "closer" than the curve y=x. So:

$\pi \int_0^1 \left[ (1-x)^2 - \left(1-\sqrt{x}\right)^2 \right] dx$

Should yield the correct result. Notice this should give you the same result you arrived at, except positive.
• February 11th 2010, 07:08 PM
But because you are rotating about y=1, you must subtract the other way around, using the radii $1-x$ and $1-\sqrt{x}$.