# Thread: Quick Surface of Revolution Question

1. ## Quick Surface of Revolution Question

Hi I have a quick question about surface of revolution. I have the problem:

Find the area of the surface obtained by rotating the curve about the x-axis.
$\displaystyle x = 1 + 2y^2$
$\displaystyle 2<=y<=3$

I know the formula for $\displaystyle S = 2\pi y ds$

When I set up the integral can I use the bounds in the same way? I.e. from 2 to 3?

2. Please define "quick" in this context. How long have you been working on it?

To answer your question directly, maybe, if that's the right thing to do. A formula does you no good if it doesn't mean anything.

3. Sorry, I was just reffering to the bounds. Im used to stuff like $\displaystyle 2<=x<=3$. Do I treat the bounds the same way (i.e. from 2 to 3) when setting up the integral? In this case $\displaystyle ds = \sqrt[]{1+(\frac{dy}{dx})^2}$

4. Ok if someone would be so kind as to check my work and show me how to simplify the following:

Rotate $\displaystyle x = 1 + 2y^2$ about the x-axis from $\displaystyle 2 <= y <= 3$

$\displaystyle y = \sqrt{\frac{x-1}{2}}$

$\displaystyle \frac{dy}{dx} = \frac{\sqrt{2}}{4\sqrt{x-1}}$

$\displaystyle ds = \sqrt{1+ (\frac{dy}{dx})^2} = \sqrt{1+ (\frac{1}{8(x-1)})}= \frac{\sqrt{8x-7}}{4}$

$\displaystyle S = 2 \pi y ds = \int_2^3 \! 2\pi (\sqrt{\frac{x-1}{2}})(\frac{\sqrt{8x-7}}{4}) \ dx.$

?