# [SOLVED]Another Solid of Revolution problem

##### Member
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
[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
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

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