91Ó°ÊÓ

An integral is given as${âˆ\neg }_{\mathrm{Ï€}}{x}^2{y}^{3}dA$

The double integration can be written as

${âˆ\neg }_{R}f(x,y)dA=\underset{\mathrm{Δ}→0}{\lim }\underset{j=1}{\overset{m}{∑}}\underset{k=1}{\overset{n}{∑}}f\left(x_j^{*},y_k^{*}\right)\mathrm{Δ}A=\underset{\mathrm{Δ}→0}{\lim }\underset{k=1}{\overset{n}{∑}}\underset{j=1}{\overset{m}{∑}}f\left(x_j^{*},y_k^{*}\right)\mathrm{Δ}A$

where

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

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

localid="1649936602006" $$\begin{gathered} {âˆ\neg }_{R}f(x,y)dA=\underset{\mathrm{Δ}→0}{\lim }\underset{j=1}{\overset{m}{∑}}\underset{k=1}{\overset{n}{∑}}(-1+j\mathrm{Δ}x{)}^2(k\mathrm{Δ}y\left)\mathrm{Δ}A \\ \underset{j=1}{\overset{m}{∑}}\underset{k=1}{\overset{n}{∑}}\right(-1+j\mathrm{Δ}x{)}^2\left(k\mathrm{Δ}y\right)\mathrm{Δ}A=\underset{j=1}{\overset{m}{∑}}(-1+j\mathrm{Δ}x{)}^2\mathrm{Δ}A\left(\underset{k=1}{\overset{n}{∑}}\left(k\mathrm{Δ}y\right)\right) \\ =\underset{j=1}{\overset{m}{∑}}(-1+j\mathrm{Δ}x{)}^2\mathrm{Δ}A\left(\mathrm{Δ}y\underset{k=1}{\overset{n}{∑}}k\right) \\ =\underset{j=1}{\overset{m}{∑}}(-1+j\mathrm{Δ}x{)}^2\mathrm{Δ}A\left(\mathrm{Δ}y\frac{n(n+1)}{2}\right) \\ =\left(\mathrm{Δ}y\frac{n(n+1)}{2}\mathrm{Δ}A\right)\left(\underset{j=1}{\overset{m}{∑}}(-1+j\mathrm{Δ}x{)}^2\right) \end{gathered}$$

$$\begin{gathered} =\left(\mathrm{Δ}y\frac{n(n+1)}{2}\mathrm{Δ}A\right)\left(\underset{j=1}{\overset{m}{∑}}\left(1-2j\mathrm{Δ}x+j^2(\mathrm{Δ}x{)}^2\right)\right) \\ =\left(\mathrm{Δ}y\frac{n(n+1)}{2}\mathrm{Δ}A\right)\left(\underset{j=1}{\overset{m}{∑}}\left(1\right)-\underset{j=1}{\overset{m}{∑}}\left(2j\mathrm{Δ}x\right)+\underset{j=1}{\overset{m}{∑}}\left(j^2(\mathrm{Δ}x{)}^2\right)\right) \\ =\left(\mathrm{Δ}y\frac{n(n+1)}{2}\mathrm{Δ}A\right)\left(\underset{j=1}{\overset{m}{∑}}\left(1\right)-2\mathrm{Δ}x\underset{j=1}{\overset{m}{∑}}\left(j\right)+(\mathrm{Δ}x{)}^2\underset{j=1}{\overset{m}{∑}}\left(j^2\right)\right) \\ =\left(\mathrm{Δ}y\frac{n(n+1)}{2}\mathrm{Δ}A\right)\left(m-2\mathrm{Δ}x\frac{m(m+1)}{2}+(\mathrm{Δ}x{)}^2\frac{m(m+1)(2m+1)}{6}\right) \\ =\left(\mathrm{Δ}y\frac{n(n+1)}{2}\right)\left(m-2\mathrm{Δ}x\frac{m(m+1)}{2}+(\mathrm{Δ}x{)}^2\frac{m(m+1)(2m+1)}{6}\right)\mathrm{Δ}A \end{gathered}$$

$$\begin{gathered} \mathrm{Δ}x=\frac{b-a}{m}=\frac{0-(-1)}{m}=\frac{1}{m} \\ \mathrm{Δ}y=\frac{d-c}{n}=\frac{2-0}{n}=\frac{2}{n} \\ \mathrm{Δ}A=\mathrm{Δ}x×\mathrm{Δ}y=\frac{1}{m}×\frac{2}{n}=\frac{2}{mn} \\ \underset{j=1}{\overset{m}{∑}}\underset{k=1}{\overset{m}{∑}}(1+j\mathrm{Δ}x{)}^2(k\mathrm{Δ}y)\mathrm{Δ}A \\ =\left(\mathrm{Δ}y\frac{n(n+1)}{2}\right)\left(m-2\mathrm{Δ}x\frac{m(m+1)}{2}+(\mathrm{Δ}x{)}^2\frac{m(m+1)(2m+1)}{6}\right)\mathrm{Δ}A \\ =\left(\left(\frac{2}{n}\right)\frac{n(n+1)}{2}\right)\left(m-2\frac{1}{m}\frac{m(m+1)}{2}+{\left(\frac{1}{m}\right)}^2\frac{m(m+1)(2m+1)}{6}\right)\left(\frac{2}{mn}\right) \\ =2\left(\frac{1}{n}\left(\frac{2}{n}\right)\frac{n(n+1)}{2}\right)\left(m\frac{1}{m}-2\frac{1}{m}\frac{m(m+1)}{2}\frac{1}{m}+{\left(\frac{1}{m}\right)}^2\frac{m(m+1)(2m+1)}{6}\frac{1}{m}\right) \\ =2\left(\frac{(n+1)}{n}\right)\left(1-2\frac{(m+1)}{m}+\frac{2}{3}\frac{(m+1)(2m+1)}{m^2}\right) \end{gathered}$$

$$\begin{gathered} {âˆ\neg }_{R}f(x,y)dA=\underset{\mathrm{Δ}→0}{\lim }\underset{j=1}{\overset{m}{∑}}\underset{k=1}{\overset{n}{∑}}(1+j\mathrm{Δ}x{)}^2(k\mathrm{Δ}y)\mathrm{Δ}A \\ =\underset{\mathrm{Δ}→0}{\lim }2\left(\frac{(n+1)}{n}\right)\left(1-2\frac{(m+1)}{m}+\frac{2}{3}\frac{(m+1)(2m+1)}{m^2}\right) \\ =\underset{m→∞}{\lim }\left(\underset{n→∞}{\lim }2\left(\frac{(n+1)}{n}\right)\left(1-2\frac{(m+1)}{m}+\frac{2}{3}\frac{(m+1)(2m+1)}{m^2}\right)\right) \\ =2\underset{m→∞}{\lim }\left(\underset{n→∞}{\lim }\left(\frac{(n+1)}{n}\right)\left(1-2\frac{(m+1)}{m}+\frac{2}{3}\frac{(m+1)(2m+1)}{m^2}\right)\right) \\ =2\underset{m→∞}{\lim }\left(1×\left(1-2\frac{(m+1)}{m}+\frac{2}{3}\frac{(m+1)(2m+1)}{m^2}\right)\right) \\ =2\left(\underset{m→∞}{\lim }1-\underset{m→∞}{\lim }2\frac{(m+1)}{m}+\underset{m→∞}{\lim }\frac{2}{3}\frac{(m+1)(2m+1)}{m^2}\right) \\ =2\left(1-2×1+\frac{2}{3}×2\right) \\ =2\left(1-2+\frac{4}{3}\right) \\ =2\left(\frac{3-6+4}{3}\right) \\ =2(\frac{1}{3}) \\ =\frac{2}{3} \end{gathered}$$