Surface Integrals in Polar Coordinates

[DrGlove]

Homework Statement

Find the area cut from the surface z = 2xy by the cylinder x^2 + y^2 = 6.
[Hint: Set up the integral using rectangular coodinates, then switch to polar coordinates.]

Homework Equations

<br /> A = \iint \sqrt{{z_x}^2+{z_y}^2+1}dxdy = \iint \sqrt{{r^2}+{r^2}{z_r}^2+{r^2}{z_\theta}^2} dr d\theta<br />

The Attempt at a Solution

<br /> <br /> \begin{gather*}<br /> z_x = 2y\\<br /> z_y = 2x\\<br /> \\<br /> A = 4\int_{0}^{\sqrt{6}} \int_{0}^{\sqrt{y^2-6}} \sqrt{4{y^2}+4{x^2}+1} dx dy\\\\<br /> A = 4\int_{0}^{\sqrt{6}} \int_{0}^{\sqrt{y^2-6}} \sqrt{4{r^2}+1} dx dy\\\\<br /> A = \int_{0}^{2\pi} \int_{0}^{\sqrt{6}} \sqrt{4{r^2}+1} dr\frac{\partial x}{\partial r} d\theta \frac{\partial y}{\partial\theta} = \int_{0}^{2\pi} \int_{0}^{\sqrt{6}} r {\cos}^2\theta \sqrt{4{r^2}+1} dr d\theta\\\\<br /> \end{gather*}<br />

From there, I can work the problem down, but I'm not sure if my conversion from dx -> dr, or dy -> d theta is even a valid operation. I don't see a ready way to do this problem if I start straight in with polar (cylindrical) coordinates, since the algebra quickly becomes very complex. I guess what I need to know, is how do I appropriately convert dx*dy -> dr*dtheta?

Any help appreciated!

 

[Dick]

The element of area dx*dy is r*dr*dtheta in polar coordinates. In a problem like this, you don't try to derive it. You just remember it and substitute.

 

[DrGlove]

D'oh! Thanks for pointing that out :P