# Surface Area

1. Mar 16, 2004

### mannaatsb

i've been stuck on this problem for an hour now. how do you find the surface area for y=(x)^1/2 when 0_<x_<2, rotating about the x axis?

2. Mar 16, 2004

### cookiemonster

You have a nice little formula for this kind of thing.

$$SA = \int_{x_i}^{x_f} 2\pi y(x) \sqrt{1 + \Big(\frac{dy}{dx}\Big)^2}\,dx$$

You'll recognize that this is $2\pi y(x)$ times the formula for the arc length. Basically what you're doing is adding up a bunch of little rings of radius y and length an infintesimal piece of the arc length. Because the radius is y, and we know that circumference = $2\pi r$, that's where the $2\pi y(x)$ comes from. The "width" of the ring depends on y and x and can be made into a triangle of base 1 and height dy/dx.

cookiemonster

Last edited: Mar 16, 2004
3. Mar 16, 2004

### mannaatsb

but how do you figure out what the dx part is? that's the part that i don't understand.

4. Mar 16, 2004

### cookiemonster

dx? dx is dx. It's necessary to evaluate the integral.

Do you mean dy/dx? You have to integrate y with respect to x to get dy/dx.

cookiemonster

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