Question

Calculate the double integral

\(\begin{aligned}{l}\int {\int\limits_R {\frac{{x{y^2}}}{{{x^2} + 1}}dA,} } \\R = \{ \left( {x,y} \right)|0 \le x \le 1, - 3 \le y \le 3\} \end{aligned}\

Short answer

Given:- function=\(\frac{{x{y^2}}}{{{x^2} + 1}},R = \{ \left( {x,y} \right)|0 \le x \le 1, - 3 \le y \le 3\} \)

To find:- Double integral

Step-by-step solution

Step (1):- (determining double integral)

Applying value of R in the integral function

Now,

\(\begin{aligned}{l}\int\limits_0^1 {\int\limits_{ - 3}^3 {\frac{{x{y^2}}}{{{x^2} + 1}}dydx} } \\\int\limits_0^1 {\frac{x}{{{x^2} + 1}}} \left( {\int\limits_{ - 3}^3 {{y^2}dy} } \right)dx\\\int\limits_0^1 {\frac{x}{{{x^2} + 1}}} \left( {\frac{{{y^3}}}{3}} \right)_{ - 3}^3dx\\\int\limits_0^1 {\frac{x}{{{x^2} + 1}}} \left( {9 + 9} \right)dx\\18\int\limits_0^1 {\frac{x}{{{x^2} + 1}}} dx\\9\int\limits_0^1 {\frac{{2x}}{{{x^2} + 1}}} dx\end{aligned}\)

Using substitution method,

Let, \({x^2} + 1 = t\)

2xdx=dt

\(\begin{aligned}{l} = 9\int\limits_0^1 {\frac{{dt}}{t}} \\ = 9\left( {{{\log }_e}t} \right)_0^1\\ = 9\left( {{{\log }_e}\left( {{x^2} + 1} \right)} \right)_0^1\\ = 9\left( {\ln \left( {{1^1} + 1} \right) - \ln \left( {0 + 1} \right)} \right)\\ = 9\ln 2 - 9\left( 0 \right)\\ = 9\ln 2\end{aligned}\)

Therefore, double integral of given function is ln2.