Surface integral and limit values $x=ar \cos(\theta), y = br\sin(\theta)$ a and b are suitably selected constants. How do I calculate surface integral for this one: $\int\int_D\ln(1+\frac{x^2}{4}+\frac{y^2}{9})dA$ when $D = $ {$ (x,y)\in \mathbf R^2$| $x \ge 0, y\ge0, 9x^2+4y^2\le36$} I have hard time to understand which are the limit values of the integral

The integral is over a quarter ellipse in the first quadrant.

It would be easier to carry out the double integral with the variable changes

$$x=2u,\>\>\>y=3v$$

Then, the ellipse becomes a unit circle

$$u^2+v^2=1$$

and the integral simplifies to

$$I=\int\int_D\ln(1+\frac{x^2}{4}+\frac{y^2}{9})dA=6\int_{Q_1}\ln(1+u^2+v^2)dudv$$

with $Q_1 = $ {$ (u,v)\in R^2$| $u \ge 0, v\ge0, u^2+v^2\le 1$}. The integral can be integrated conveniently with the polar coordinates,

$$I=6\int_0^{\pi/2}\int_0^1 \ln(1+r^2)rdrd\theta=6\pi\ln2$$