First construct the region [latex]D[/latex] as a Type I region (Figure 5). Here [latex]{D} = {\left \{{(x,y)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {2,{\frac{1}{2}x}} \ {\leq} \ y \ {\leq} \ {1} \right \}}[/latex]. Then we have

[latex]\large{\underset{D}{\displaystyle\iint}{x^2}{e^{xy}}{dA} = \displaystyle\int_{x=0}^{x=2}\displaystyle\int_{y=1/2x}^{y=1}x^2e^{xy}dydx.}[/latex]

Figure 5. We can express region [latex]\small{D}[/latex] as a Type I region and integrate from [latex]\small{y=\frac{1}{2}x}[/latex] to [latex]\small{y=1}[/latex], between the lines [latex]\small{x=0}[/latex] and [latex]\small{x=2}[/latex].

Therefore, we have
[latex]\begin{align}\displaystyle\int_{x=0}^{x=2}\displaystyle\int_{y=\frac12x}^{y=1}x^2e^{xy}dydx &=\int_{x=0}^{x=2}[\int_{y=\frac12x}^{y=1}x^2e^{xy}dydx] &\text{Iterated for a Type I region.} \\&=\int_{x=0}^{x=2}[x^2\frac{e^xy}x]\Bigg|_{y=\frac12x}^{y=1}dx &\text{Integrate with respect to }y\text{ using }u\text{-substitution} \\& &\text{ with }u=xy\text{ where }x\text{ is held constant} \\&=\int_{x=0}^{x=2}[xe^x-xe^{\frac{x^2}2}]dx &\text{Integrate with respect to }x\text{ using }u\text{-substitution with}u=\frac12x^2 \\&=[xe^x-e^x-e^{\frac12x^2}]\bigg|_{x=0}^{x=2}=2\end{align}[/latex]

Notice that [latex]D[/latex] can be seen as either a Type I or a Type II region, as shown in Figure 7. However, in this case describing [latex]D[/latex] as Type I is more complicated than describing it as Type II. Therefore, we use [latex]D[/latex] as a Type II region for the integration.

Figure 7. The region [latex]D[/latex] in this example can be either (a) Type I or (b) Type II.

Choosing this order of integration, we have[latex]\begin{align}\underset{D}{\displaystyle\iint}(3x^2+y^2)dA&=\displaystyle\int_{y=-2}^{y=3}\displaystyle\int_{x=y^2-3}^{x=y+3}(3x^2+y^2)dxdy &\quad &\text{Interated integral, Type II region}.\\&=\displaystyle\int_{y=-2}^{y=3}(x^2+xy^2\Bigg|_{y^2-3}^{y+3}dy &\quad &\text{Integrate with respect to }x. \\&=\displaystyle\int_{y=-2}^{y=3}\left((y+3)^2+(y+3)y^2-(y^2-3)^3-(y^2-3)y^2\right)dy \\&=\displaystyle\int_{-2}^{3}(54+27y-12y^2+2y^3+8y^4-y^6)dy &\quad &\text{Integrate with respect to }y. \\&=\left[54y+\frac{27y^2}2-4y^3+\frac{y^4}2+\frac{8y^5}5-\frac{y^7}7\right]\bigg|_{-2}^3 \\&=\frac{2,375}7.\end{align}[/latex]

The region [latex]D[/latex] is not easy to decompose into any one type; it is actually a combination of different types. So we can write it as a union of three regions [latex]D_1[/latex], [latex]D_2[/latex], and [latex]D_3[/latex] where,

[latex]{D_1} = {\left \{{(x,y)}{\mid}{-2} \ {\leq} \ {x} \ {\leq} \ {0,0} \ {\leq} \ {y} \ {\leq} \ {(x+y)^2} \right \}}[/latex],

[latex]{D_2} = {\left \{{(x,y)}{\mid}{0} \ {\leq} \ {y} \ {\leq} \ {4,0} \ {\leq} \ {x} \ {\leq} \ {(y-{\frac{1}{16}y^3})} \right \}}[/latex].

These regions are illustrated more clearly in Figure 9.

Figure 9. Breaking the region into three subregions makes it easier to set up the integration.

Here [latex]D_1[/latex] is Type I and [latex]D_2[/latex] and [latex]D_3[/latex] are both of Type II. Hence,

[latex]\begin{align} \underset{D}{\displaystyle\iint}(2x+5y)dA&=\underset{D_1}{\displaystyle\iint}(2x+5y)dA+\underset{D_2}{\displaystyle\iint}(2x+5y)dA+\underset{D_3}{\displaystyle\iint}(2x+5y)dA \\ &=\displaystyle\int_{x=-2}^{x=0}\displaystyle\int_{y=0}^{y=(x+2)^2}(2x+5y)dydx+\displaystyle\int_{y=0}^{y=4}\displaystyle\int_{x=0}^{x=y-(1/16)y^3}(2x+5y)dxdy+\displaystyle\int_{y=-4}^{y=0}\displaystyle\int_{x=-2}^{x=y-(1/16)y^3}(2x+5y)dxdy \\ &=\displaystyle\int_{x=-2}^{x=0}\left[\frac12(2+x)^2(20+24x+5x^2)\right] \\ &\,\,\,+\displaystyle\int_{y=0}^{y=4}\left[\frac1{256}y^6-\frac7{16}y^4+6y^2\right] +\displaystyle\int_{y=-4}^{y=0}\left[\frac1{256}y^6-\frac7{16}y^4+6y^2+10y-4\right] \\ &=\frac{40}3+\frac{1,664}{35}-\frac{1,696}{35}=\frac{1,304}{105}. \end{align}[/latex]

The region as presented is of Type I. To reverse the order of integration, we must first express the region as Type II. Refer to Figure 10.

Figure 10. Converting a region from Type I to Type II.

We can see from the limits of integration that the region is bounded above by [latex]y=2-x^{2}[/latex] and below by [latex]y=0[/latex], where [latex]x[/latex] is in the interval [latex][0,{\sqrt{2}}][/latex]. By reversing the order, we have the region bounded on the left by [latex]x=0[/latex] and on the right by [latex]{x} = {{\sqrt{2-y}}}[/latex] where [latex]y[/latex] is in the interval [latex][0,2][/latex]. We solved [latex]y=2-x^2[/latex] in terms of [latex]x[/latex] to obtain [latex]{x} = {{\sqrt{2-y}}}[/latex].

Hence

[latex]\begin{align} \displaystyle\int_0^{\sqrt2}\displaystyle\int_0^{2-x^2}xe^{x^2}dydx&=\displaystyle\int_0^2\displaystyle\int_0^{\sqrt{2-y}}xe^{x^2}dxdy \,\,\,\,\,\text{ Reverse the order of integration then use substitution} \\ &=\displaystyle\int_0^2\left[\frac12ex^2\bigg|_0^{\sqrt{2-y}}\right]dy=\displaystyle\int_0^2\frac12(e^{2-y}-1)dy=-\frac12(e^{2-y}+y)\bigg|_0^2 \\ &=\frac12(e^2-3). \end{align}[/latex].

A sketch of the region appears in Figure 11.

Figure 11. A triangular region [latex]R[/latex] for integrating in two ways.

We can complete this integration in two different ways.

a. One way to look at it is by first integrating [latex]y[/latex] from [latex]y=0[/latex] to [latex]y=1-x[/latex] vertically and then integrating [latex]x[/latex] from [latex]x=0[/latex]         to [latex]x=1[/latex]:

[latex]\hspace{5cm}\begin{align} \underset{R}{\displaystyle\iint}f(x,y)dxdy&=\displaystyle\int_{x=0}^{x=1}\displaystyle\int_{y=0}^{y=1-x}(x-2y)dydx=\displaystyle\int_{x=0}^{x=1}[xy-2y^2]_{y=0}^{y=1-x}dx \\ &=\displaystyle\int_{x=0}^{x=1}\left[x(1-x)-(1-x)^2\right]dx=\displaystyle\int_{x=0}^{x=1}[-1+3x-2x^2]dx \\ &=\left[-x+\frac32x^2-\frac23x^3\right]_{x=0}^{x=1}=-\frac16 \end{align}[/latex]

b. The other way to do this problem is by first integrating [latex]x[/latex] from [latex]x=0[/latex] to [latex]x=1-y[/latex] horizontally and then integrating [latex]y[/latex] from [latex]y=0[/latex] to [latex]y=1[/latex]:

[latex]\hspace{5cm}\begin{align} \underset{R}{\displaystyle\iint}f(x,y)dxdy&=\displaystyle\int_{y=0}^{y=1}\displaystyle\int_{x=0}^{x=1-y}(x-2y)dxdy=\displaystyle\int_{y=0}^{y=1}[\frac12x^2-2xy]_{x=0}^{x=1-y}dy \\ &=\displaystyle\int_{y=0}^{y=1}\left[\frac12(1-y)^2-2y(1-y)\right]dy=\displaystyle\int_{y=0}^{y=1}[\frac12-3y+\frac52y^2]dy \\ &=\left[\frac12y-\frac32y^2+\frac56y^3\right]_{y=0}^{y=1}=-\frac16 \end{align}[/latex]