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

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.

• The mathematical term for $G$ is the unit disk, not the unit circle. Its boundary is the unit circle. – Jack Lee Jun 28 at 22:05

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.

• Could you explain why the integrals involving $sin$ $x$ and $cos$ $x$ are zero? I know that they are $2\pi$ periodic, but isn't the power interfering here? – Gabi G Jun 28 at 18:15
• @GabiG It is an odd power. – A.Γ. Jun 28 at 18:20

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.

$$\int_{-\pi}^{\pi} \int_0^1 (r^{2018}cos(x)^{2017} - r^{2018}sin(x)^{2017})drdx=\frac{1}{2019}\int_{-\pi}^{\pi}(cos^{2017}x-sin^{2017}x) dx =0$$ 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.

• So is the answer $0$ or $\frac{1}{2019}$? – Gabi G Jun 28 at 18:14
• 0, $\frac{1}{2019}$ is the result of integrating $\int_0^1r^{2018}$ alone – ryszard eggink Jun 28 at 18:15