Polar Coordinates and Double Integrals Problem 1:
Find the area enclosed by the ellipse $\displaystyle \frac {1} {r} = 1 – 0.6 \cos(\theta)$.
We know $0\leq \theta\leq 2\pi$.
We know $0\leq r\leq 1/(1-0.6\cos(\theta))$.
Questions:
What I’m not sure about it how to set up the double integral. What am i integrating? And when I sketched the picture the ellipse went through $(2.5, 0)$ and $(-0.5, 0)$ and $(0, 1)$ and $(0, -1)$. Is that useful? 
Similar problem:
Given $x = ar\cos(\theta)$ and $y = br\sin(\theta)$, find the area enclosed by the curve: 
$(x^2/a^2 + y^2/b^2)^2 = xy/c^2$
Again, I can find the limits but I’m not sure how to set this up correctly. 
 A: If you do want to do this with a double integral, the answer to your question "what am I integrating" is $1$. But unless you explicitly have to do that, you can save yourself some work. An ellipse is a stretched circle -- a unit circle stretched by a factor of the two semi-axes in two perpendicular directions. Thus its area is just the area of the unit circle, $\pi$, times the stretching factors, which are the semi-axes. So all you have to do is to find the semi-axes.
For the first problem, you can see in your sketch (or from the fact that $r$ is an even function of $\theta$) that the semi-axes are parallel to the coordinate axes. The horizontal one is immediate from Matthew's comment: $a= (2.5+0.625)/2=\frac{5}{4}+\frac{5}{16}=\frac{25}{16}=\left(\frac{5}{4}\right)^2$. For the vertical one, you have to maximize the vertical coordinate:
$$r\sin\theta = \frac{\sin\theta}{1-0.6\cos\theta}\to\max$$
$$\left(\frac{\sin\theta}{1-0.6\cos\theta}\right)'=\frac{\cos\theta(1-0.6\cos\theta)-0.6\sin^2\theta}{(1-0.6\cos\theta)^2}=\frac{\cos\theta-0.6}{(1-0.6\cos\theta)^2}=0$$
$$\cos\theta=0.6$$
$$\frac{\sin\theta}{1-0.6\cos\theta}=\frac{\sqrt{1-0.6^2}}{1-0.6^2}=\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{0.8}=\frac{5}{4}$$
Thus, the area of the ellipse is $\left(\frac{5}{4}\right)^2\frac{5}{4}\pi=\left(\frac{5}{4}\right)^3\pi$.
