# Surface Area Double Integral Problem

Find the area of the portion of the plane $z = x + 3y$ that lies inside the elliptical cylinder with equation $\frac{x^2}{4}+\frac{y^2}{9} = 1.$

The formula for finding this is $\int \int_{R} \sqrt{1+z_x^2+z_y^2}dxdy$, however, in practice I know that it is easier to convert to polar coordinates. The problem that I'm having is that $\frac{x^2}{4}+\frac{y^2}{9} = 1$ doesn't convert very smoothly. I get:

$\frac{r^2cos^2\theta}{4}+\frac{r^2sin^2\theta}{9} = 1$.

Typically, what I see with these questions is a simple way to take advantage of the fact that $cos^2\theta + sin^2\theta = 1.$ This time, however, the fractions prevent that. As a result, everything I attempt ends up giving me EXTREMELY convoluted integrals that I'm sure aren't the intended methods. What should I be doing here to set the integrals up?

You can turn your ellipse into a circle as you change variables, letting $u=\frac{x}{2}$ and $v=\frac{y}{3}$ and then integrate as usual. Don't forget the jacobian! If you alreaady know the formula of the area of an ellipse, then it's easier, since the the are will just be $\int\int\limits_{\frac{x^{2}}{4}+\frac{y^{2}}{4}\leq 1} \sqrt{1+1+9} dx dy=\sqrt{11 } \times$area of the ellipse = $6\pi \sqrt{11}$