I am not sure how to work this one out.
I am suppose to find the area of this parametric equation.
$$y = b\sin\theta, x = a\cos\theta$$
$$0 \leq 0 \leq 2\pi$$
I set up the equation in the memorized formula.
$$\int_0^{2\pi} \sqrt{1 + \left(\frac{b\cos\theta}{-a\sin\theta}\right)^2}d\theta$$
$$\int_0^{2\pi} \sqrt{1 + \frac{b^2\cos^2\theta}{a^2\sin^2\theta}}d\theta$$
$$\int_0^{2\pi} \sqrt{1 + \frac{b^2}{a^2}\cdot \csc^2\theta}d\theta$$
From here I am at a loss of what to do, I tried some trig subsitution but I do not think that will work and it only seems to complicate the problem.
Best Answer
Hint: scale $x$ and $y$ suitably and you get a circle. What does that do to the area?