1. ## Integral Question

Two parts to this problem:
1. Show that the area of the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\displaystyle {\pi}ab$;

2. Find the volume enclosed by the ellipsoid $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ by integrating the area of a horizontal cross section.

I can't find any similar examples in my textbook and I am really stumped on this one, I'm taking a guess that this is meant to be solved by double integrals? If so, I am terrible at setting those up >.< If someone could point me in the right direction for solving it would help a lot =)

2. Originally Posted by Em Yeu Anh
Two parts to this problem:
1. Show that the area of the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\displaystyle {\pi}ab$;
Parameterize the ellipse by $\displaystyle x=a\cos(\theta),y=b\sin(\theta)$ and remember the area for a parametric curve is $\displaystyle \int_a^b x'(t)y(t)dt$. From this we see that $\displaystyle ab\int_0^{2\pi}\sin^2(\theta)d\theta=\pi ab$

Follow suit for the second one.

EDIT: Typo.

3. Originally Posted by Drexel28
Parameterize the ellipse by $\displaystyle x=a\cos(\theta),y=b\sin(\theta)$ and remember the area for a parametric curve is $\displaystyle \int_a^b x'(t)y(t)dt$. From this we see that $\displaystyle ab\int_0^{2\pi}\sin^2(\theta)d\theta=\pi ab$

Follow suit for the second one.

EDIT: Typo.
Thanks a bunch.
so $\displaystyle x'(t) = -asint$ is that correct?
Would the area for the parametric curve then be $\displaystyle -ab\int_0^{2{\pi}}sin^2{\theta}d{\theta}$?