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.