I'm having trouble visualizing the solution to this double integral of an ellipse
Solution 1:
Before that change of variable, the region over which you are integrating is the ellipse $D$. But$$(x,y)\in D\iff x=3+3r\cos\theta\wedge y=-4+4r\sin\theta\text{ for some $(\theta,r)\in[0,2\pi]\times[0,1]$},$$and therefore after the change of variables the region over which you will be integrating is the rectangle $[0,2\pi]\times[0,1]$.