Compute $\int^2_0\int^2_y \frac{28}{3}(x^2+xy)dxdy$

Question

\begin{align*} \int^2_0\int^2_y \frac{28}{3}(x^2+xy)dxdy&=\frac{28}{3}\int^2_0\int^2_y x^2dxdy+\frac{28}{3}\int^2_0\int^2_y xydxdy\\ &=\frac{28}{3}\int^2_0\left[\frac{x^3}{3}\right]^2_ydx+\frac{28}{3}\times \int^2_0y \int^2_yx dxdy\\\ &=\frac{28}{3}\int^2_0\left(\frac{8-y^3}{3}\right)dy+\frac{28}{3} \times \int^2_0 y \times \left[\frac{x^2}{2}\right]^2_y dx \\ &=\frac{28}{3}\times\frac{1}{3}\int^2_08dy-\int^2_0y^3dy+\frac{28}{3}\times \int^2_0 y\left(2-\frac{y^2}{2}\right)dy\\ &=\frac{28}{3}\times\frac{1}{3}\times\left[8y\right]^2_0-\left[\frac{y^4}{4}\right]^2_0+\frac{28}{3}\times\int^2_0 y\times2\left(1-\frac{\frac{y^2}{2}}{2}\right)dy\\ &=\frac{28}{3}\times\frac{1}{3}(16-4)+\frac{28}{3}\times2\times \int^2_0 y\times\left(1-\frac{\frac{y^2}{2}}{2}\right)dy\\ &=\frac{28}{3}\times\frac{12}{3}+\frac{28}{3}\times2\times\int^2_0y\times\left(1-\frac{y^3}{4}\right)dy\\ &=\frac{28}{3}\times4+\frac{28}{3}\times2\times\int^2_0ydy-\frac{y^3}{4}dy\\ &=\frac{28}{3}\times4+\frac{28}{3}\times2\times\int^2_0y-\frac{1}{4}\times\int^2_0y^3dy\\ &=\frac{28}{3}\times4+\frac{28}{3}\times2\times \left[\frac{y^2}{2}\right]^2_0-\frac{1}{4}\times\left[\frac{y^4}{4}\right]^2_0\\ &=\frac{28}{3}\times4+\frac{28}{3}\times2\left(\frac{4}{2}-\frac{1}{4}\times\frac{16}{4}\right)\\ &=\frac{28}{3}\times4+\frac{28}{3}\times2\left(2-1\right)\\ &=\frac{28}{3}\times4+\frac{28}{3}\times2\\ &=56\\ \end{align*}

I double-checked the solution in wolfram alpha and it's a match. However, would this be the best method in deriving the solution? Is there a more succinct method?

Answer 1

Your way is fine, maybe we can simplify by a different notation and steps as follows

$$\int^2_0\int^2_y \frac{28}{3}(x^2+xy)dxdy=\frac{28}{3}\int^2_0 \left[\frac{x^3}3+\frac{x^2y}2\right]_y^2dy=\frac{28}{3}\int^2_0\frac{8}3+\frac{4y}2 -\frac{y^3}3-\frac{y^3}2dy=$$

$$=\frac{28}{3}\int^2_0\frac{8}3+2y -\frac{5y^3}6dy=\frac{28}{3}\left[\frac{8}3y+y^2-\frac{5y^4}{24}\right]_0^2=\frac{28}{3}\left(\frac{16}3+4-\frac{80}{24}\right)=56$$

As an alternative, we can proceed with the equivalent

$$\int^2_0\int^x_0 \frac{28}{3}(x^2+xy)dydx=\frac{28}{3}\int^2_0 \left(x^3+\frac {x^3}2\right)dx=\frac{28}{3}\left[\frac {3x^4}8\right]_0^2=\frac{28}{3}\cdot 6=56$$

Answer 2

Swapping $x$ and $y$ first gives a slightly simpler intermediate step:

$$\begin{align*} \int_0^2 \int_y^2 \frac{28}3 (x^2+xy) \, dx \, dy &= \frac{28}3 \int_0^2 \int_{\color{red}0}^x (x^2+xy) \, dy \, dx \\ &= \frac{28}3 \int_0^2 \left(x^2y + \frac{xy^2}2\right)\bigg|_{y=x} - \left(x^2y + \frac{xy^2}2\right)\bigg|_{y=0} \, dx \\ &= \frac{28}3 \int_0^2 \frac{3x^3}2 \, dx \end{align*}$$