0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.