5.1 Double integrals over rectangular regions  (Page 5/12)

[link] shows how the calculation works in two different ways.

1. First integrate with respect to y and then integrate with respect to x :
   
   $\begin{array}{cc}\hfill {\displaystyle \underset{R}{\iint }f(x,y)dA}& ={\displaystyle {\int }_{x=0}^{x=2}{\displaystyle {\int }_{y=0}^{y=3}(3x^2-y)dydx}}\hfill \\ & ={\displaystyle {\int }_{x=0}^{x=2}\left({\displaystyle \underset{y=0}{\overset{y=3}{\int }}(3x^2-y)dy}\right)}dx={\displaystyle {\int }_{x=0}^{x=2}\left[{3x^{2}y-\frac{y^2}{2}|}_{y=0}^{y=3}\right]dx}\hfill \\ & ={\displaystyle {\int }_{x=0}^{x=2}\left(9x^2-\frac{9}{2}\right)}dx={3x^3-\frac{9}{2}x|}_{x=0}^{x=2}=15.\hfill \end{array}$
2. First integrate with respect to x and then integrate with respect to y :
   
   $\begin{array}{cc}\hfill {\displaystyle \underset{R}{\iint }f(x,y)dA}& ={\displaystyle {\int }_{y=0}^{y=3}{\displaystyle {\int }_{x=0}^{x=2}(3x^2-y)dxdy}}\hfill \\ & ={\displaystyle {\int }_{y=0}^{y=3}\left({\displaystyle {\int }_{x=0}^{x=2}(3x^2-y)dx}\right)}dy={\displaystyle {\int }_{y=0}^{y=3}\left[{x^3-xy|}_{x=0}^{x=2}\right]dy}\hfill \\ & ={\displaystyle {\int }_{y=0}^{y=3}\left(8-2y\right)dy}={8y-y^2|}_{y=0}^{y=3}=15.\hfill \end{array}$

1. We can express ${\displaystyle \underset{R}{\iint }x\text{sin}(xy)dA}$ in the following two ways: first by integrating with respect to $y$ and then with respect to $x;$ second by integrating with respect to $x$ and then with respect to $y.$
   
   $\begin{array}{}\\ \\ \\ \\ {\displaystyle \underset{R}{\iint }x\text{sin}(xy)dA}\hfill & & & \\ ={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}{\displaystyle \underset{y=1}{\overset{y=2}{\int }}x\text{sin}(xy)dydx}}\hfill & & & \text{Integrate first with respect to}y.\hfill \\ ={\displaystyle \underset{y=1}{\overset{y=2}{\int }}{\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}x\text{sin}(xy)dxdy}}\hfill & & & \text{Integrate first with respect to}x.\hfill \end{array}$
2. If we want to integrate with respect to y first and then integrate with respect to $x,$ we see that we can use the substitution $u=xy,$ which gives $du=xdy.$ Hence the inner integral is simply ${\displaystyle \int \text{sin}u}du$ and we can change the limits to be functions of x ,
   
   ${\displaystyle \underset{R}{\iint }x\text{sin}(xy)dA}={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}{\displaystyle \underset{y=1}{\overset{y=2}{\int }}x\text{sin}(xy)dydx}}={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}\left[{\displaystyle \underset{u=x}{\overset{u=2x}{\int }}\text{sin}(u)du}\right]}dx.$
   
   However, integrating with respect to $x$ first and then integrating with respect to $y$ requires integration by parts for the inner integral, with $u=x$ and $dv=\text{sin}(xy)dx.$
   Then $du=dx$ and $v=-\frac{\text{cos}(xy)}{y},$ so
   
   ${\displaystyle \underset{R}{\iint }x\text{sin}(xy)dA}={\displaystyle \underset{y=1}{\overset{y=2}{\int }}{\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}x\text{sin}(xy)dxdy}}={\displaystyle \underset{y=1}{\overset{y=2}{\int }}\left[-{\frac{x\text{cos}(xy)}{y}|}_{x=0}^{x=\pi }+\frac{1}{y}{\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}\text{cos}(xy)dx}\right]}dy.$
   
   Since the evaluation is getting complicated, we will only do the computation that is easier to do, which is clearly the first method.
3. Evaluate the double integral using the easier way.
   
   $\begin{array}{cc}\hfill {\displaystyle \underset{R}{\iint }x\text{sin}(xy)dA}& ={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}{\displaystyle \underset{y=1}{\overset{y=2}{\int }}x\text{sin}(xy)dydx}}\hfill \\ & ={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}\left[{\displaystyle \underset{u=x}{\overset{u=2x}{\int }}\text{sin}(u)du}\right]}dx={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}\left[{\text{−}\text{cos}u|}_{u=x}^{u=2x}\right]}dx={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}\left(\text{−}\text{cos}2x+\text{cos}x\right)dx}\hfill \\ & ={-\frac{1}{2}\text{sin}2x+\text{sin}x|}_{x=0}^{x=\pi }=0.\hfill \end{array}$