## Problem on a line integral over a circle

Let $C$ be a circle of radius $a$ centered at the origin, traversed counterclockwise. For what nonzero value of $a$ is $\oint_C \bigl(-y + \frac{1}{3} y^3 + x^2 y \bigr) dx = 0$?
• ## Solution

Recall that

#### Selecting an integration method

Our initial approach to solve the line integral is by direct parameterization
A parameterization of the circle of radius $a$ is $$x(\theta ) = a \cos \theta, \qquad y(\theta) = a \sin \theta.$$
Hence, $$dx = -a \sin \theta d \theta, \qquad dy = a \cos \theta d\theta.$$
The integral gives rise to terms like $x^2 y \ dx = -a^4 \cos^2 \theta \ \sin^2 \theta \ d\theta$.
Terms like this can be integrated, but they can be messy. We will try to find another method of integration that is easier.
Our second approach is to solve the integral using the fundamental theorem of line integrals.
Recall that
We identify $u(x,y) = -y + \frac{1}{3}y^3 +x^2 y$ and $v(x,y) = 0$.
Because $\partial_y u \neq \partial_x v$, the vector field is not conservative. Hence, we can not use the fundamental theorem.
Our third approach is to solve the integral using Green's theorem.

#### Solution using Green's Theorem

Recall that
We note that $C$ is closed because it is a circle. The vector field is a polynomial in $x,y$, so it is well behaved inside $C$.
Recall Green's Theorem
We identify $L(x,y) = -y + \frac{1}{3} y^3 + x^2 y$ and $M(x,y) = 0$.
We compute the partial derivatives \begin{align}
\partial_y L(x,y) &= -1 + y^2 + x^2\\
\partial_x M(x,y) &= 0
\end{align}
Applying the theorem and computing the partial derivatives, we see that $$\oint_C \bigl(-y + \frac{1}{3} y^3 + x^2 y \bigr) dx = \iint_D (1 - y^2 -x^2) dxdy$$ where D is the disc of radius $a$.
We wish to find the value of $a$ such that $\frac{\pi}{2} (2 a^2 - a^4) = 0$
Dividing by constants and $a^2$, we get $2 =a^2$.
The radius $a$ for which the line integral is zero is $a= \sqrt{2}$.