Question

Evaluate the double integral \({\int\limits_D {\int y } ^2}dA,\,D = \left\{ {\left( {x,y} \right)\left| { - 1 \le y \le 1,\, - y - 2 \le x \le y} \right|} \right\}\).

Short answer

Value of integral is \(\frac{4}{3}\).

Step-by-step solution

Step 1:

\( \Rightarrow \int {{y^2}} dA = \int\limits_{ - 1}^1 {\int\limits_{ - y - 2}^y {{y^2}} } dxdy\)

Integrating with respect to \(x\).

\(I = \int_{ - 1}^1 {\int_{ - y - 2}^1 {{y^2}} } dxdy\)

\(I = \int\limits_{ - 1}^1 {{y^2}\left( {y - ( - y - 2)} \right)} dy\)

\( = \int_{ - 1}^1 {{y^2}} (y + y + 2)dy\)

\( = \int_{ - 1}^1 {{y^2}} (2y + 2)dy\)

\(I = \int_{ - 1}^1 {\left( {2{y^3} + 2{y^2}} \right)} dy\)

Now, \(2{y^3}\) and \(2\left( { - {y^3}} \right)\) are opposite and equal.

\(I = \int_{ - 1}^1 2 {y^2}dy \)

\( = \frac{2}{3}\left( {{y^3}} \right)_{ - 1}^1\)

\( = \frac{2}{3}(1 + 1)\)

\( = \frac{4}{3}\)

Hence, Value of integral is \(\frac{4}{3}\).