calculate the double integral. ∬r 3xy²/(x² + 1) da, r = {(x, y)|0 ≤ x ≤ 3, -2 ≤ y ≤ 2}

Answer:

$24\ln(10)$

Explanation:

Step1: Set up iterated integral

$$\int_{0}^{3}\int_{-2}^{2}\frac{3xy^{2}}{x^{2}+1}dydx$$

Step2: Integrate with respect to $y$ first

$$\int_{0}^{3}\frac{3x}{x^{2}+1}\left[\frac{y^{3}}{3}\right]{-2}^{2}dx=\int{0}^{3}\frac{3x}{x^{2}+1}\left(\frac{8}{3}-\left(-\frac{8}{3}\right)\right)dx=\int_{0}^{3}\frac{16x}{x^{2}+1}dx$$

Step3: Use substitution

Let $u = x^{2}+1$, then $du = 2xdx$. When $x = 0$, $u = 1$; when $x = 3$, $u = 10$. So the integral becomes $8\int_{1}^{10}\frac{du}{u}$.

Step4: Integrate with respect to $u$

$$8[\ln(u)]_{1}^{10}=8(\ln(10)-\ln(1)) = 8\ln(10)=24\ln(10)$$