- #1

DrGlove

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## 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

[tex]

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

[/tex]

## The Attempt at a Solution

[tex]

\begin{gather*}

z_x = 2y\\

z_y = 2x\\

\\

A = 4\int_{0}^{\sqrt{6}} \int_{0}^{\sqrt{y^2-6}} \sqrt{4{y^2}+4{x^2}+1} dx dy\\\\

A = 4\int_{0}^{\sqrt{6}} \int_{0}^{\sqrt{y^2-6}} \sqrt{4{r^2}+1} dx dy\\\\

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\\\\

\end{gather*}

[/tex]

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!

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