I am trying to understand a problem of integration - which does not look so complicated but I really struggle with a variable change. Here it is: $$ D=\{(x,y)\in R^2:4(x-1)^2+9(y-1)^2<36,y>1,x-y>0\}$$ calculate $$\iint_D xydxdy $$.
The domain $D$ is an ellipse centred in $(1,1)$ which can be parametrized as: $$x=1+3r\cos\theta,y=1+2r\sin\theta$$ which is the change of variable to be done. The jacobian for this change of variable is $6r$ and the domain can be described as: $$D=\{(1+3r\cos\theta,1+2r\sin\theta):0<r<1,0<\theta<\arctan\left(\frac{3}{2}\right)\}$$ hence the integral becomes: $$6\int_0^{\arctan(3/2)}\int_0^1r+3r^2\cos\theta+2r^2\sin\theta+3r^3\sin(2\theta)drd\theta $$
My question is this: I really don't understand the upper limit in $\theta=\arctan(3/4)$ : why is it not $\pi/4$ ? Seems to me that this should be the good angle as the "left/up" side part of the domain is limited by the line $x=y$ which has a $\theta=\pi/4$ angle with horizontal lines... I am clearly missing something basic here but cannot figure out what... Any help would be very much appreciated !
Marc
EDIT: just to clarify, I do understand that 3/2 comes from writing that $x=y$ on the ellipse boundary which, under the change of variables, means $1+3r\cos\theta=1+2r\sin\theta\Rightarrow\frac{\sin\theta}{\cos\theta}=\tan\theta=\frac{3}{2}$, but the point is that with my eyes I see $\pi/4$... so there clearly is a property of change of variables I don't understand...