# Set up double integral of ellipse in polar coordinates?

How do you set up a double integral for an ellipse in polar coordinates without using Jacobian or Greens Theorem?

I can’t seem to figure out what (or if) the limits of r can possible be.

$x = a\cos(t), y = b\cos(t), ( z = 0)$

$x^2/a + y^2/b = 1$

Thank you.

#### Solutions Collecting From Web of "Set up double integral of ellipse in polar coordinates?"

Try using elliptical coordinates:
$$\begin{cases} x = \sqrt{a}\,r\cos t \\ y = \sqrt{b}\,r \sin t\end{cases}.$$

Note that $x^2/a + y^2/b = r$, so the ellipse itself is given by $r = 1$. To get the whole interior, let $0 < r < 1$. Also note that $t$ has to vary between $0$ and $2\pi$ to cover the whole ellipse (even though $t$ does not quite measure the angle between the radius vector of $(x,y)$ and the $x$-axis, as it does in polar coordinates).

You might want to try following through on GEdgar’s suggestion. First, verify that

$$r=\frac{b^2}{a-\sqrt{a^2-b^2}\cos\,t}$$

is the polar equation of an ellipse with semiaxes $a$, and $b$, with the origin as one of the foci. You can then use the formula

$$\int_0^{2\pi}\frac{r^2}{2}\mathrm d\theta$$

(which is what you’d obtain if you set up the double integral and evaluate before substituting in $r$). In particular, you can simplify things by noting that for this case,

$$\int_0^{2\pi}\frac{r^2}{2}\mathrm d\theta=\int_0^\pi r^2\mathrm d\theta$$

(why?)