Surface Area of y=(x)^1/2 Rotating About the x Axis

[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?

 

[cookiemonster]

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

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

You'll recognize that this is [itex]2\pi y(x)[/itex] 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 = [itex]2\pi r[/itex], that's where the [itex]2\pi y(x)[/itex] 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

 

[mannaatsb]

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

 

[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