Problem 56 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Evaluate the integral \(\iint_{R} \frac{x - y}{x^{2}+y^{2}+1} d A\), 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 1: 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 \(0 \leq r \leq 1\) and \(0 \leq \theta \leq 2\pi\). Next, we set up the integral in polar coordinates:

Step 2: Set Up the Integral

The integral to evaluate is: $$ \int_{0}^{2\pi} \int_{0}^{1} \frac{r\cos\theta - r\sin\theta}{r^{2}+1} \cdot r \, dr d\theta $$ Now, we observe symmetry:

Step 3: 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:

Step 4: Write the Result

Since the integral evaluates to zero due to symmetry, we have: $$ \iint_{R} \frac{x - y}{x^{2}+y^{2}+1} d A = 0 $$