# Surface area when revolved around something other than the x-axis

Consider the region R bounded by f(x)=x^3, y =1, and x = 2. Find the surface area when f(x) is revolved around the line x=5.

I know that the equation for surface area is the intergal from a to b of 2pi*r*h dx
in all of the examples that I had done in class, everything was around the x or y axis. I don't even understand how to get my r or my h in this case. If anyone has any suggestions, it would be greatly appreciated.

## Answers and Replies

Consider the region R bounded by f(x)=x^3, y =1, and x = 2. Find the surface area when f(x) is revolved around the line x=5.

I know that the equation for surface area is the intergal from a to b of 2pi*r*h dx
in all of the examples that I had done in class, everything was around the x or y axis. I don't even understand how to get my r or my h in this case. If anyone has any suggestions, it would be greatly appreciated.

Do you understand where the formula you've mentioned comes from (when it rotates around y-axis)?

Here the case is similar, except, the radius of the cylindrical shell is different. The height of the shell is still the same. Can you work out the radius?

I do vaguely remember deriving and showing where the formula came from in class. My problem is that I just can't visualize and understand how to alter it to make it fit the way we managed to in the examples around the y or x axis for the radius.

Try drawing a diagram.

You have an ALMOST trapezium-like element with sides on x-axis and x = 2, y = 1 and y = x^3 from 0 to 2. Now the formula you know is:

$$V=2\pi \int_a^b xf(x)dx$$

Here, x is the radius, when it rotates about the y-axis. Now it rotates about x = 5. How would you define the radius now?

5-x? I think

Exactly. Plugging that into your formula should give you the correct result.

ooo I get it! Ok thank you soooo much