Question

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

localid="1649936867482" ${\iint }_x{x}^{2}ydA$

where$R=\left\{\right(x,y)\mid 1\le x\le 3\text{and}0\le y\le 2\}$

Short answer

The value of integral is$\frac{2}{3}$

Step-by-step solution

Step 1. Given information

An integral is given as${\iint }_{\pi }{x}^2{y}^{3}dA$

Step 2. Evaluating integral

The double integration can be written as

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

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 $\left(x_j^{*},y_i^{*}\right)$choose points $\left(x_j,y_k\right)=(-1+j\Delta x,0+k\Delta y)=(-1+j\Delta x,k\Delta y)$ for each j and k.

localid="1649936602006" $$\begin{gathered} {\iint }_{R}f(x,y)dA=\underset{\Delta \to 0}{\lim }\sum _{j=1}^m\sum _{k=1}^n(-1+j\Delta x{)}^2(k\Delta y\left)\Delta A \\ \sum _{j=1}^m\sum _{k=1}^n\right(-1+j\Delta x{)}^2\left(k\Delta y\right)\Delta A=\sum _{j=1}^m(-1+j\Delta x{)}^2\Delta A\left(\sum _{k=1}^n\left(k\Delta y\right)\right) \\ =\sum _{j=1}^m(-1+j\Delta x{)}^2\Delta A\left(\Delta y\sum _{k=1}^{n}k\right) \\ =\sum _{j=1}^m(-1+j\Delta x{)}^2\Delta A\left(\Delta y\frac{n(n+1)}{2}\right) \\ =\left(\Delta y\frac{n(n+1)}{2}\Delta A\right)\left(\sum _{j=1}^m(-1+j\Delta x{)}^2\right) \end{gathered}$$

$$\begin{gathered} =\left(\Delta y\frac{n(n+1)}{2}\Delta A\right)\left(\sum _{j=1}^m\left(1-2j\Delta x+j^2(\Delta x{)}^2\right)\right) \\ =\left(\Delta y\frac{n(n+1)}{2}\Delta A\right)\left(\sum _{j=1}^m\left(1\right)-\sum _{j=1}^m\left(2j\Delta x\right)+\sum _{j=1}^m\left(j^2(\Delta x{)}^2\right)\right) \\ =\left(\Delta y\frac{n(n+1)}{2}\Delta A\right)\left(\sum _{j=1}^m\left(1\right)-2\Delta x\sum _{j=1}^m\left(j\right)+(\Delta x{)}^2\sum _{j=1}^m\left(j^2\right)\right) \\ =\left(\Delta y\frac{n(n+1)}{2}\Delta A\right)\left(m-2\Delta x\frac{m(m+1)}{2}+(\Delta x{)}^2\frac{m(m+1)(2m+1)}{6}\right) \\ =\left(\Delta y\frac{n(n+1)}{2}\right)\left(m-2\Delta x\frac{m(m+1)}{2}+(\Delta x{)}^2\frac{m(m+1)(2m+1)}{6}\right)\Delta A \end{gathered}$$

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

$$\begin{gathered} {\iint }_{R}f(x,y)dA=\underset{\Delta \to 0}{\lim }\sum _{j=1}^m\sum _{k=1}^n(1+j\Delta x{)}^2(k\Delta y)\Delta A \\ =\underset{\Delta \to 0}{\lim }2\left(\frac{(n+1)}{n}\right)\left(1-2\frac{(m+1)}{m}+\frac{2}{3}\frac{(m+1)(2m+1)}{m^2}\right) \\ =\underset{m\to \infty }{\lim }\left(\underset{n\to \infty }{\lim }2\left(\frac{(n+1)}{n}\right)\left(1-2\frac{(m+1)}{m}+\frac{2}{3}\frac{(m+1)(2m+1)}{m^2}\right)\right) \\ =2\underset{m\to \infty }{\lim }\left(\underset{n\to \infty }{\lim }\left(\frac{(n+1)}{n}\right)\left(1-2\frac{(m+1)}{m}+\frac{2}{3}\frac{(m+1)(2m+1)}{m^2}\right)\right) \\ =2\underset{m\to \infty }{\lim }\left(1\times \left(1-2\frac{(m+1)}{m}+\frac{2}{3}\frac{(m+1)(2m+1)}{m^2}\right)\right) \\ =2\left(\underset{m\to \infty }{\lim }1-\underset{m\to \infty }{\lim }2\frac{(m+1)}{m}+\underset{m\to \infty }{\lim }\frac{2}{3}\frac{(m+1)(2m+1)}{m^2}\right) \\ =2\left(1-2\times 1+\frac{2}{3}\times 2\right) \\ =2\left(1-2+\frac{4}{3}\right) \\ =2\left(\frac{3-6+4}{3}\right) \\ =2(\frac{1}{3}) \\ =\frac{2}{3} \end{gathered}$$