Find \( \iint_{R} \frac{x y^{2}}{x^{2}+1} d A, R=[0,2] \times[-3,3] \) Round your answer to four decimal places.

We start with the double integral \[ \iint_R \frac{x y^2}{x^2 + 1}\, dA, \] where the region is \[ R = [0,2] \times [-3,3]. \] ### Step 1. Write the integral as an iterated integral Since the region is rectangular, the integral can be written as \[ \int_{x=0}^{2} \int_{y=-3}^{3} \frac{x y^2}{x^2+1} \,dy\,dx. \] ### Step 2. Evaluate the inner integral (with respect to \( y \)) For a fixed \( x \), the integrand is \[ \frac{x y^2}{x^2+1}, \] where \( \frac{x}{x^2+1} \) is a constant in terms of \( y \). Thus, we have \[ \int_{y=-3}^{3} \frac{x y^2}{x^2+1} dy = \frac{x}{x^2+1} \int_{y=-3}^{3} y^2\,dy. \] Notice that \( y^2 \) is an even function. Therefore, \[ \int_{y=-3}^{3} y^2\,dy = 2\int_{0}^{3} y^2\,dy. \] Compute the integral: \[ \int_{0}^{3} y^2\,dy = \left[\frac{y^3}{3}\right]_{0}^{3} = \frac{27}{3} = 9. \] Thus, \[ \int_{y=-3}^{3} y^2\,dy = 2 \times 9 = 18. \] So the inner integral becomes \[ \frac{x}{x^2+1} \times 18 = \frac{18x}{x^2+1}. \] ### Step 3. Evaluate the outer integral (with respect to \( x \)) Now we need to compute \[ \int_{x=0}^{2} \frac{18x}{x^2+1}\,dx. \] Factor out the constant: \[ 18 \int_{0}^{2} \frac{x}{x^2+1}\,dx. \] ### Step 4. Use substitution for the \( x \)-integral Let \[ u = x^2+1 \quad \text{so that} \quad du = 2x\,dx. \] Then, \[ \frac{du}{2} = x\,dx. \] Change the limits: - When \( x = 0 \), \( u = 0^2+1 = 1 \). - When \( x = 2 \), \( u = 2^2+1 = 5 \). Substitute into the integral: \[ 18 \int_{u=1}^{5} \frac{1}{u}\cdot\frac{du}{2} = 18 \cdot \frac{1}{2} \int_{1}^{5} \frac{du}{u} = 9 \int_{1}^{5} \frac{du}{u}. \] ### Step 5. Compute the logarithmic integral The integral is \[ \int_{1}^{5} \frac{du}{u} = \ln|u|\Big|_{1}^{5} = \ln(5) - \ln(1). \] Since \( \ln(1) = 0 \), we have \[ \int_{1}^{5} \frac{du}{u} = \ln(5). \] Thus, the overall value becomes \[ 9\ln(5). \] ### Step 6. Final answer The value of the given double integral is \[ 9\ln(5) \approx 14.4849, \] rounded to four decimal places.