Another Solid of Revolution problem

[MacLaddy1]

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]

[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]

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

 

[MacLaddy1]

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