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[SOLVED] Another Solid of Revolution problem

MacLaddy

Member
Jan 29, 2012
52
I'm finding myself stuck again. This time it is more in the set up then the solving.

Find the volume of the following solid of revolution.

The region bounded by \(\displaystyle y=\frac{1}{x^2}\), y=0, x=2, x=8, and revolved about the y-axis.

I am trying to use the shell method to solve this, as it seems the best scenario in this situation. This is my set-up,

\(\displaystyle 2\pi\int_2^8 (radius)(height)dx\)
\(\displaystyle 2\pi\int_2^8 (x+2)(\frac{1}{x^2})dx\)
\(\displaystyle 2\pi\int_2^8 (\frac{1}{x}+\frac{2}{x^2})dx\)

Now here is where I am getting into a snag. Following this through I am coming up with a completely different solution to the manual, and the solution manual is showing the first integration step as \(\displaystyle 2\pi\int_2^8 (\frac{1}{x})dx\), but I can't figure out how they are coming up with that integral. It is showing a final solution of \(\displaystyle \pi\ln(16)\)

Any help, or a kick in the right direction, would be greatly appreciated.

Mac
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
[snip]
Now here is where I am getting into a snag. Following this through I am coming up with a completely different solution to the manual, and the solution manual is showing the first integration step as \(\displaystyle 2\pi\int_2^8 (\frac{1}{x})dx\), but I can't figure out how they are coming up with that integral. It is showing a final solution of \(\displaystyle \pi\ln(16)\)

Any help, or a kick in the right direction, would be greatly appreciated.

Mac
[/snip]
That's because the radius of your solid of revolution is just $x$, not $x+2$. $x$ is always measured from the $y$-axis (i.e. $x=0$), so there's no need to add 2 to this value; in this case, $x$ just happens to be between 2 and 8.

On the other hand, the radius would be $x+2$ if you were revolving the region about the line $x=-2$.

I hope this clarifies things!
 

CaptainBlack

Well-known member
Jan 26, 2012
890
I'm finding myself stuck again. This time it is more in the set up then the solving.

Find the volume of the following solid of revolution.

The region bounded by \(\displaystyle y=\frac{1}{x^2}\), y=0, x=2, x=8, and revolved about the y-axis.

I am trying to use the shell method to solve this, as it seems the best scenario in this situation. This is my set-up,

\(\displaystyle 2\pi\int_2^8 (radius)(height)dx\)
\(\displaystyle 2\pi\int_2^8 (x+2)(\frac{1}{x^2})dx\)
\(\displaystyle 2\pi\int_2^8 (\frac{1}{x}+\frac{2}{x^2})dx\)

Now here is where I am getting into a snag. Following this through I am coming up with a completely different solution to the manual, and the solution manual is showing the first integration step as \(\displaystyle 2\pi\int_2^8 (\frac{1}{x})dx\), but I can't figure out how they are coming up with that integral. It is showing a final solution of \(\displaystyle \pi\ln(16)\)

Any help, or a kick in the right direction, would be greatly appreciated.

Mac
Where did a radius of \(x+2\) come from? Why not \(x\)?

CB
 

MacLaddy

Member
Jan 29, 2012
52
Thanks Chris and Captain for the replies.

I had a hunch that this is where my misunderstanding was. So when using the shell method, and it is rotated directly among one of the axis', then the radius can be considered x (or y), and the start and stop point is the limits for my integration?

That explanation probably isn't very good, but I think I understand. When using the disc or washer method then the radius is found more similarly to what I was doing? I think I'm getting confused between methods here. I probably need more practice.

Thanks guys.

Mac