Question

Evaluate the following integrals using the method of your choice. A sketch is helpful. ${\iint }_R\frac{x-y}{x^2+y^2+1}dA;R$ is the region bounded by the unit circle centered at the origin.

Short answer

Question: Evaluate the integral ${\iint }_R\frac{x-y}{x^2+y^2+1}dA$, where $R$ is the region bounded by the unit circle centered at the origin. Answer: The integral evaluates to 0 due to symmetry in the region and the integrand.

Step-by-step solution

Convert to Polar Coordinates

We have $x=r\cos \theta$ and $y=r\sin \theta$. Also, the Jacobian for polar coordinates is r, so $dA=rd(r)d(\theta )$. The integrand becomes $\frac{r\cos \theta -r\sin \theta }{r^2+1}$. The region of integration will be $0\le r\le 1$ and $0\le \theta \le 2\pi$. Next, we set up the integral in polar coordinates:

Set Up the Integral

The integral to evaluate is: $${\int }_0^{2\pi }{\int }_0^1\frac{r\cos \theta -r\sin \theta }{r^2+1}\cdot rdrd\theta$$ Now, we observe symmetry:

Observe Symmetry

The region $R$ is symmetric about both the x-axis and y-axis. Also, the integrand $\frac{r\cos \theta -r\sin \theta }{r^2+1}$ is odd with respect to y-axis, as when we replace $\theta$ with $\pi -\theta$, the integrand becomes its negation. Thus, the integral evaluates to 0 over the symmetric region. Finally, we have the answer:

Write the Result

Since the integral evaluates to zero due to symmetry, we have: $${\iint }_R\frac{x-y}{x^2+y^2+1}dA=0$$