1. ## Double integral

I'm asked to calculate the double integral

$\displaystyle \int{\int_{D}(x+y)^2}dxdy$ with $\displaystyle D=\left\{(x,y)\in \mathbb{R}^2: (x-2y)^2+(3x+3y)^2\leq 9\right\}$

It seems the best to me, to calculate this with a substitution and adjusted domain S:

Take $\displaystyle u=x-2y, v=3x+3y$ so that our new domain $\displaystyle S$ will be $\displaystyle \left\{(u,v):u^2+v^2\leq 9\right\}$

and we have $\displaystyle x=\frac{1}{9}(3u+2v), y=\frac{1}{9}(v-3u)$

And our new integral becomes $\displaystyle \frac{1}{9}\int\int_{S}v^2dS$. However I wasn't sure I calculated the Area element $\displaystyle dS$ correctly.

I found $\displaystyle dS= 9dudv$. Is this all correct?

2. Originally Posted by Dinkydoe
I'm asked to calculate the double integral

$\displaystyle \int{\int_{D}(x+y)^2}dxdy$ with $\displaystyle D=\left\{(x,y)\in \mathbb{R}^2: (x-2y)^2+(3x+3y)^2\leq 9\right\}$

It seems the best to me, to calculate this with a substitution and adjusted domain S:

Take $\displaystyle u=x-2y, v=3x+3y$ so that our new domain $\displaystyle S$ will be $\displaystyle \left\{(u,v):u^2+v^2\leq 9\right\}$

and we have $\displaystyle x=\frac{1}{9}(3u+2v), y=\frac{1}{9}(v-3u)$

And our new integral becomes $\displaystyle \frac{1}{9}\int\int_{S}v^2dS$. However I wasn't sure I calculated the Area element $\displaystyle dS$ correctly.

I found $\displaystyle dS= 9dudv$. Is this all correct?
In 2D space the change of variables means that we now have

$\displaystyle \int{\int_{D}(g(u,v))}det|d(x,y)/d(u,v)|dudv$

So take the determinent. This means we want

|x1,x2/y1,y2|dudv = 1/9

So you are correct.