How to find the area enclosed by the ellipse $b^2x^2 + a^2y^2=a^2b^2$? I need to find the area enclosed by the ellipse $b^2x^2 + a^2y^2=a^2b^2$, and I know it involves taking the integral, but I'm not sure what function I should be taking the integral of or how to find the upper and lower bounds.
Also I realize this is the standard equation for an ellipse if you divide by $a^2b^2$ on both sides.
 A: Let's do an area preserving ($\det(T)=1$) change of coordinates, shall we:
We want a transformation which is something like $(x,y) \to (ax,by)$.  If we had such a transformation, then, plugging in, we'd get something like: $b^2a^2x^2 + a^2b^2y^2=a^2b^2 \iff x^2+y^2=1$ which is just the equation of a circle.
So our matrix should be $\begin{bmatrix} a & 0 \\ 0 & b\end{bmatrix}$ because
$$\begin{bmatrix} a & 0 \\ 0 & b\end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} ax \\ by \end{bmatrix}$$
Notice, though, that this is not an area-preserving transformation unless $ab$ happens to equal $1$.  However, this one is:
$$\left(\frac 1{\sqrt{ab}}\begin{bmatrix} a & 0 \\ 0 & b\end{bmatrix}\right)\begin{bmatrix} x \\ y \end{bmatrix} = \frac 1{\sqrt{ab}}\begin{bmatrix} ax \\ by \end{bmatrix}$$
So doing the change of coordinates $(x,y) \to (\sqrt{\frac ab}x, \sqrt{\frac ba}y)$ won't change the area of your ellipse.
Plugging this in, we get $x^2 + y^2 = ab$.  I assume you can find the area of this?
A: You need to start by solving for $y$,
$$\begin{eqnarray} \frac{b^2}{a^2} x^2 + y^2 &= b^2 \\
y^2 &= b^2 - \frac{b^2}{a^2} x^2 \\
y &= \pm b\sqrt{ 1 -  \frac{x^2}{a^2} } \end{eqnarray}$$
Now, notice that you've got some symmetry there, and so the area you want is
$$2b \int_{-a}^a \sqrt{ 1 -  \frac{x^2}{a^2} }\, \mathrm{d}x.$$
To solve it, I suggest a trig substitution.
A: I assume that you are familiar to some basic concepts in multiple variable calculus.
We will make use of the new coordinates: $x=ar\cos(\theta),\; y=br\sin(\theta)$. Now the area $A$ of the ellipse is given by:
\begin{equation}
A=\int_{0}^{2\pi}d\theta\int_{0}^1 abr\; dr=\pi a b
\end{equation}
The $abr$ in the integral comes from evaluating the determinant of the Jacobian matrix: http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant.
A: Here the solution with all the steps!
http://en.wikipedia.org/wiki/Ellipse
