Surface integral and limit values

by engineerstudent   Last Updated September 11, 2019 13:20 PM

$x=ar \cos(\theta), y = br\sin(\theta)$ 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



Answers 1


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

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

which can be integrated with the polar coordinates,

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

Quanto
Quanto
September 11, 2019 13:15 PM

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