# Area of a Surface of Revolution

1. Jul 26, 2006

### steelphantom

I'm having more Calculus troubles here. Here's the problem:

Write the definite integral that represents the area of the surface formed by revolving the graph of $$f(x) = 81 - x^2$$ on the interval $$[0, 9]$$ about the y axis; evaluate the integral to determine the surface area.

By knowing that $$f(x) = 81 - x^2$$, $$f'(x) = -2x$$. I then set up the integral for surface area, and I get this:

$$2\pi \int_{0}^{9} (81 - x^2)\sqrt{1 + (-2x)^2} dx$$

Assuming I did that correctly, I can't figure out for the life of me how to evaluate the integral. Any ideas?

Edit: I'm pretty sure I set up the integral wrong. I think it should be:

$$2\pi \int_{0}^{9} (x)\sqrt{1 + (-2x)^2} dx$$

If so, I think I found the answer, and it comes out to be about 3068.

Last edited: Jul 26, 2006
2. Jul 26, 2006

Do you know why the first is not?

3. Jul 27, 2006

### steelphantom

Well, the forumula for surface area, as far as I know, is

$$2\pi \int_{a}^{b} r(x)\sqrt{1 + f'(x)} dx$$

where r(x) is the radius of the ring at the given x. 81 - x^2 isn't the radius; x is. Is that the correct reasoning?

4. Jul 27, 2006

### end3r7

Yep, that's the reason. :)