Question

Use Definition $13.4$to evaluate the double integrals in Exercises $29–32$.

${\iint }_{R}xydA$

where

$R=\left\{\right(x,y)\mid 0\le x\le 2\&1\le y\le 4\}$

Short answer

The value of integral is$15$

Step-by-step solution

Step 1. Given information

An integral is given as${\iint }_{R}xydA$

Step 2. Evaluating integral

By definition the double integration is

${\iint }_{R}f(x,y)dA=\underset{\Delta \to 0}{\lim }\sum _{j=1}^m\sum _{k=1}^{n}f\left(x_j^{*},y_k^{*}\right)\Delta A=\underset{\Delta \to 0}{\lim }\sum _{k=1}^n\sum _{j=1}^{m}f\left(x_j^{*},y_k^{*}\right)\Delta A$

where

$$\begin{gathered} x_j=a+j\Delta x \\ {y}_k=b+k\Delta y \\ \Delta A=\Delta x\times \Delta y \\ \Delta x=\frac{b-a}{m} \\ \Delta y=\frac{d-c}{n} \end{gathered}$$

The starred points $\left(x_j^{*},y_k^{*}\right)$lets choose $\left(x_j,y_k\right)=(0+j\Delta x,1+k\Delta y)=(j\Delta x,1+k\Delta y)$ for each jand k

$$\begin{gathered} {\iint }_{R}f(x,y)dA=\underset{\Delta \to 0}{\lim }\sum _{j=1}^m\sum _{k=1}^n\left(j\Delta x\right)(1+k\Delta y)\Delta A \\ \sum _{j=1}^m\sum _{k=1}^n\left(j\Delta x\right)(1+k\Delta y)\Delta A=\sum _{j=1}^m\left(j\Delta x\right)\Delta A\left(\sum _{k=1}^n(1+k\Delta y)\right) \\ =\sum _{j=1}^m\left(j\Delta x\right)\Delta A\left(\sum _{k=1}^{n}1+\sum _{k=1}^{n}k\Delta y\right) \\ =\sum _{j=1}^m\left(j\Delta x\right)\Delta A\left(n+\Delta y\frac{n(n+1)}{2}\right) \\ =\left(n+\Delta y\frac{n(n+1)}{2}\right)\Delta A\left(\sum _{j=1}^m\left(j\Delta x\right)\right) \\ =\left(n+\Delta y\frac{n(n+1)}{2}\right)\Delta A\left(\Delta x\sum _{j=1}^m\left(j\right)\right) \\ =\left(n+\Delta y\frac{n(n+1)}{2}\right)\left(\Delta x\frac{m(m+1)}{2}\right)\Delta A \end{gathered}$$

Recall,

$$\begin{gathered} \Delta x=\frac{b-a}{m}=\frac{2-0}{m}=\frac{2}{m} \\ \Delta y=\frac{d-c}{n}=\frac{4-1}{n}=\frac{3}{n} \\ \Delta A=\Delta x\times \Delta y=\frac{2}{m}\times \frac{3}{n}=\frac{6}{mn} \\ \sum _{j=1}^m\sum _{k=1}^n\left(j\Delta x\right)(1+k\Delta y)\Delta A=\left(n+\Delta y\frac{n(n+1)}{2}\right)\left(\Delta x\frac{m(m+1)}{2}\right)\Delta A \\ =\left(n+\frac{3}{n}\frac{n(n+1)}{2}\right)\left(\frac{2}{m}\frac{m(m+1)}{2}\right)\left(\frac{6}{mn}\right) \\ =6\left(n\frac{1}{n}+\frac{3}{n}\frac{1}{n}\frac{n(n+1)}{2}\right)\left(\frac{1}{m}\frac{2}{m}\frac{m(m+1)}{2}\right) \\ =6\left(1+\frac{3}{2}\frac{(n+1)}{n}\right)\left(\frac{(m+1)}{m}\right) \end{gathered}$$

The double integration is

$$\begin{gathered} {\iint }_{R}f(x,y)dA=\underset{\Delta \to 0}{\lim }\sum _{j=1}^m\sum _{k=1}^n\left(j\Delta x\right)(1+k\Delta y)\Delta A \\ =\underset{\Delta \to 0}{\lim }6\left(1+\frac{3}{2}\frac{(n+1)}{n}\right)\left(\frac{(m+1)}{m}\right) \\ =\underset{m\to \infty }{\lim }\left(\underset{n\to \infty }{\lim }6\left(1+\frac{3}{2}\frac{(n+1)}{n}\right)\left(\frac{(m+1)}{m}\right)\right) \\ =6\underset{m\to \infty }{\lim }\left(\underset{n\to \infty }{\lim }\left(1+\frac{3}{2}\frac{(n+1)}{n}\right)\left(\frac{(m+1)}{m}\right)\right) \\ =6\underset{m\to \infty }{\lim }\left(\underset{n\to \infty }{\lim }1+\underset{n\to \infty }{\lim }\left(\frac{3}{2}\frac{(n+1)}{n}\right)\left(\frac{(m+1)}{m}\right)\right) \\ =6\underset{m\to \infty }{\lim }\left(\left(1+\frac{3}{2}\right)\times \left(\frac{(m+1)}{m}\right)\right) \\ =6\left(1+\frac{3}{2}\right)\left(\underset{m\to \infty }{\lim }\frac{(m+1)}{m}\right) \\ =6\left(\frac{5}{2}\right)\left(1\right) \\ =15 \end{gathered}$$