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

1. Feb 21, 2010

### BoldKnight399

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.

2. Feb 21, 2010

### kbaumen

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?

3. Feb 21, 2010

### BoldKnight399

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.

4. Feb 21, 2010

### kbaumen

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. Feb 21, 2010

### BoldKnight399

5-x? I think

6. Feb 21, 2010

### kbaumen

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

7. Feb 21, 2010

### BoldKnight399

ooo I get it! Ok thank you soooo much