# Green's Theorem: Properly parameterizing in polar coordinates

My professor provided our class the folloring example problem demonstrating Green's Theorem:

Evaluate $\int F \cdot dr$ where $C$ is the circle with radius $2$ oriented clockwise and $F$ is the function $$F = (3y+\frac{e^{80/x^2}}{x^{2017}},-4x-\frac{e^{2017}}{e^{2018}})$$

The problem is easy when Green's Theorem is applied, as the nasty exponents cancel out and you are left taking the integral of $7$ ($-7$ if the circle were counter-clockwise). However, what he did next is confusing me.

Since it is a circle, the integral should be in polar coordinates, thus the integral should be $7\iint r dr d\theta$. However, my professor evaluated it as simply $7\iint dA$ and did not add an $r$. Then, when he finished integrating $r$ from $0$ to $2$ and $\theta$ from $0$ to $2\pi$, his answer is $7\cdot 4\pi$, which is $28\pi$.

When I attempted the problem, I did add the $r$ and ended up with $24\pi$.

Am I not supposed to add the extra polar coordinate components in this problem? Or did my professor make a mistake? Below is a picture of his example.

## 1 Answer

$$7 \int_0^{2\pi} \int_0^2 r \,\,dr d\theta=7 \cdot (2\pi)\frac{2^2}{2}=28\pi$$