${\int }_0^3{\int }_0^{2 x}\left(x^2\+y^2\right) \mathit{\ⅆ}y \mathit{\ⅆ}x\=\frac{189}{2}$ 

${\int }_0^6{\int }_{y\/2}^3\left(x^2\+y^2\right) \mathit{\ⅆ}x \mathit{\ⅆ}y\=\frac{189}{2}$ 

Evaluate the given integral

${\int }_0^3{\int }_0^{2 x}\left(x^2\+y^2\right) \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{189}{2}$$$

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻ 

Table 5.4.1(a)   Integration in the order $\mathrm{dy} \mathrm{dx}$ 

Guided by Table 5.4.1(a), reverse the order of integration

${\int }_0^6{\int }_{y\/2}^3\left(x^2\+y^2\right) \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{189}{2}$$$

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻ 

Table 5.4.1(b)   Integration in the order $\mathrm{dx} \mathrm{dy}$

Evaluate the given integral

$\mathrm{int}\left(x^2\+y^2\,\left[y\=0..2 x\,x\=0..3\right]\right)$ = $\frac{189}{2}$$$

Graph the region of integration

$\mathrm{plots}:-\mathrm{inequal}\left(\left\{x\ge 0\,x\le 3\,y\ge 0\,y\le 2 x\right\}\,x\=0..3\,y\=0..6\right)$

Reverse the order of integration

$q≔\mathrm{Int}\left(x^2\+y^2\,\left[x\=y\/2..3\,y\=0..6\right]\right)$

${\∫}_0^6{\∫}_{\frac{1}{2}y}^3\left(x^2\+y^2\right)\ⅆx\ⅆy$

$\mathrm{value}\left(q\right)$ = $\frac{189}{2}$$$