calculate $\int_\gamma (y^{2018} + y^2e^{xy^2})dx + (x^{2018} + 2xye^{xy^2})dy$ when $\gamma$ is the unit circle.

Question

Let $G$ be the unit circle $\{x^2+y^2 <1\}$. Let $\gamma$ be the boundary of G, meaning $\{x^2+y^2 =1\}$ in a positive orientation.

I need to calculate $\int_\gamma \omega$ when $\omega = (y^{2018} + y^2e^{xy^2})dx + (x^{2018} + 2xye^{xy^2})dy$.

I checked and determined that $\omega$ is not exact, so I've decided to use Green's theorem here. So, I get: $$\int_\gamma \omega = \int_G (2018x^{2017} - 2018y^{2017})dxdy$$ and after I change variables to polar coordinates I get: $\int_0^{2\pi} \int_0^1 (r^{2018}\cos(\theta)^{2017} - r^{2018}\sin(\theta)^{2017})drd\theta $

However, I am having a problem with solving this integral.

Help would be appreciated.

Answer 1

You are almost done.

$$\int_0^{2\pi} \int_0^1 (r^{2018}cos(\theta)^{2017} - r^{2018}sin(\theta)^{2017})drd\theta=$$

$$\int_0^{2\pi} \int_0^1 (r^{2018}cos(\theta)^{2017}-\int_0^{2\pi} \int_0^1 r^{2018}sin(\theta)^{2017})drd\theta=$$

$$\int_0^{2\pi}cos(\theta)^{2017}d\theta \int_0^1 r^{2018}dr-\int_0^{2\pi}sin(\theta)^{2017}d\theta \int_0^1 r^{2018}dr=0$$

Note that the integrals involving $\sin (\theta)$ and $\cos (\theta)$ are zero.

Therefore the final answer is zero.

Answer 2

$\int_{-\pi}^{\pi} \int_0^1 (r^{2018}cos(x)^{2017} - r^{2018}sin(x)^{2017})drdx=\frac{1}{2019}\int_{-\pi}^{\pi}(cos^{2017}x-sin^{2017}x) dx =0$ It is easy to see that for all odd $r$ $\int_{-\pi}^{\pi}sin^rx=0, \int_{-\pi}^{\pi}cos^rx=0 $, because $sin^{r}(x)=-sin^{r}(y)$ where $x\in[-\pi,0]$ and $y\in[0,\pi]$ and similar argument with $cosx$, but you have to pick the intervals more cleverly. Also it's obvious that integrals are convergent as both $sin,cosx\leq1$. The first equality is from Fubini.

Answer 3

The integral that you are working with after applying Green's Theorem is especially suited for symmetry about the origin. You can think about it this way: the expression $f(x,y) =2018(x^{2017} - y^{2017})$ has the property that $f(x,y) = -f(-x, -y)$. If there is a perfect one-to-one correspondence of the points $(x,y)$ and $(-x, -y)$ in our region, then the integral must be zero. I'll leave it to you to explain why our region has a one-to-one correspondence of these points.