Evaluating a contour-integral.

Question

Consider the ellipse $C$ given by $x^2 + y^2/4 = 1$. How to evaluate

$$\int_C x^2 \, \nu(d(x,y))$$

where $\nu$ is the Lebesgue length measure on $C$?

I am not sure if this can be computed like a usual contour integral in complex analysis?

Answer 1

Choose a parametrization (any will do), for instance $x(t) = \cos t, y(t) = 2 \sin t$. From the statement I assume $C$ to be the whole ellipse (not just a segment of it), which means that $t \in [0, 2 \pi]$.

Your integral will now be

$$\int \limits _0 ^{2 \pi} x(t) ^2 \sqrt {x'(t) ^2 + y' (t) ^2} \Bbb d t = \\ \int \limits _0 ^{2 \pi} \cos ^2 t \sqrt {\sin ^2 t + 4 \cos ^2 t} \Bbb d t = \\ \int \limits _0 ^{2 \pi} \cos ^2 t \sqrt {1 + 3 \cos ^2 t} \Bbb d t ,$$

which to my knowledge cannot be expressed in terms of elementary functions (you will need to use elliptic functions).