Problem 84 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\iint_{R} \frac{x y}{1+x^{2}+y^{2}} d A ; R=\\{(x, y): 0 \leq y \leq x, 0 \leq x \leq 2\\}$$

Step-by-Step Solution

Answer

Question: Evaluate the double integral given by the region R, where $$\int \int_R \frac{x y}{1+x^{2}+y^{2}} dA$$, and R is the region bounded by $$0 \leq y \leq x$$ and $$0 \leq x \leq 2$$. Answer: \(\frac{\pi}{4}\)

Step 1: Identify the limits of integration

Based on the given region description, we can identify the limits of integration for both variables: $$0 \leq y \leq x$$ $$0 \leq x \leq 2$$

Step 2: Set up the double integral

Now that we have the limits of integration, we can set up the double integral: $$\int_{0}^{2} \int_{0}^{x} \frac{x y}{1+x^{2}+y^{2}} dy dx$$

Step 3: Evaluate the inner integral

First, we need to integrate with respect to y: $$\int_{0}^{2} \left[ \frac{1}{2} \arctan \left( \frac{y}{x} \right) \right]_{0}^{x} dx = \int_{0}^{2} \left( \frac{1}{2} \arctan(1) - 0 \right) dx$$ The term in the square brackets simplifies to: $$\frac{1}{2} \arctan(1) = \frac{1}{2} \left( \frac{\pi}{4} \right) = \frac{\pi}{8}$$

Step 4: Evaluate the outer integral

Now we will integrate with respect to x: $$\int_{0}^{2} \frac{\pi}{8} dx = \left[ \frac{\pi}{8} x \right]_{0}^{2} = \frac{\pi}{8} (2) - \frac{\pi}{8}(0) = \frac{\pi}{4}$$ The value of the double integral is \(\frac{\pi}{4}\).