91Ó°ÊÓ

An integral is given as${âˆ\neg }_{R}x{y}^{3}dA$

Use identity 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

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

For starred points $\left(x_j^{*},y_k^{*}\right)$let's choose $\left(x_j,y_k\right)=(-2+j\mathrm{Δ}x,-1+k\mathrm{Δ}y)$ for each jand k

working on the Riemann sum gives,

$$\begin{gathered} \underset{j=1}{\overset{m}{∑}}\underset{k=1}{\overset{n}{∑}}(-2+j\mathrm{Δ}x)(-1+k\mathrm{Δ}y{)}^3\mathrm{Δ}A=\underset{j=1}{\overset{m}{∑}}(-2+j\mathrm{Δ}x)\mathrm{Δ}A\left(\underset{k=1}{\overset{n}{∑}}(-1+k\mathrm{Δ}y{)}^3\right) \\ =\underset{j=1}{\overset{m}{∑}}(-2+j\mathrm{Δ}x\left)\mathrm{Δ}A\left(\underset{i=1}{\overset{n}{∑}}(k\mathrm{Δ}y-1{)}^3\right) \\ \underset{k=1}{\overset{n}{∑}}\right(k\mathrm{Δ}y-1{)}^3=\underset{i=1}{\overset{n}{∑}}\left((k\mathrm{Δ}y{)}^3-3(k\mathrm{Δ}y{)}^{2}1+3k\mathrm{Δ}y(1{)}^2-\left(1^3\right)\right) \\ =\underset{k=1}{\overset{n}{∑}}\left(k^3\mathrm{Δ}{y}^3-3k^2\mathrm{Δ}{y}^2+3k\mathrm{Δ}y-1\right) \\ =\underset{k=1}{\overset{n}{∑}}{k}^3\mathrm{Δ}{y}^3-\underset{k=1}{\overset{n}{∑}}3k^2\mathrm{Δ}{y}^2+\underset{k=1}{\overset{n}{∑}}3k\mathrm{Δ}y-\underset{k=1}{\overset{n}{∑}}\left(1\right) \\ =\mathrm{Δ}{y}^3\underset{k=1}{\overset{n}{∑}}{k}^3-3\mathrm{Δ}{y}^2\underset{k=1}{\overset{n}{∑}}{k}^2+3\mathrm{Δ}y\underset{k=1}{\overset{n}{∑}}k-\underset{k=1}{\overset{n}{∑}}\left(1\right) \\ =\mathrm{Δ}{y}^3\frac{n^2(n+1{)}^2}{4}-3\mathrm{Δ}{y}^2\frac{n(n+1)(2n+1)}{6}+3\mathrm{Δ}y\frac{n(n+1)}{2}-n \\ =\mathrm{Δ}{y}^3\frac{n^2(n+1{)}^2}{4}-\mathrm{Δ}{y}^2\frac{n(n+1)(2n+1)}{2}+\mathrm{Δ}y\frac{3n(n+1)}{2}-n \end{gathered}$$

Hence,

$$\begin{gathered} \underset{j=1}{\overset{m}{∑}}\underset{k=1}{\overset{n}{∑}}(-2+j\mathrm{Δ}x)(-1+k\mathrm{Δ}y{)}^3\mathrm{Δ}A= \\ =\underset{j=1}{\overset{m}{∑}}(-2+j\mathrm{Δ}x)\mathrm{Δ}A\left(\mathrm{Δ}{y}^3\frac{n^2(n+1{)}^2}{4}-\mathrm{Δ}{y}^2\frac{n(n+1)(2n+1)}{2}+\mathrm{Δ}y\frac{3n(n+1)}{2}-n\right) \\ =\mathrm{Δ}A\left(\mathrm{Δ}{y}^3\frac{n^2(n+1{)}^2}{4}-\mathrm{Δ}{y}^2\frac{n(n+1)(2n+1)}{2}+\mathrm{Δ}y\frac{3n(n+1)}{2}-n\right)\underset{j=1}{\overset{m}{∑}}(-2+j\mathrm{Δ}x) \\ =\mathrm{Δ}A\left(\mathrm{Δ}{y}^3\frac{n^2(n+1{)}^2}{4}-\mathrm{Δ}{y}^2\frac{n(n+1)(2n+1)}{2}+\mathrm{Δ}y\frac{3n(n+1)}{2}-n\right)\left(\underset{j=1}{\overset{m}{∑}}(-2)+\underset{j=1}{\overset{m}{∑}}\left(j\mathrm{Δ}x\right)\right) \\ =\mathrm{Δ}A\left(\mathrm{Δ}{y}^3\frac{n^2(n+1{)}^2}{4}-\mathrm{Δ}{y}^2\frac{n(n+1)(2n+1)}{2}+\mathrm{Δ}y\frac{3n(n+1)}{2}-n\right)\left(-2m+\mathrm{Δ}r\frac{m(m+1)}{2}\right) \end{gathered}$$

Recall,

$$\begin{gathered} \mathrm{Δ}x=\frac{b-a}{m}=\frac{2-(-2)}{m}=\frac{4}{m} \\ \mathrm{Δ}y=\frac{d-c}{n}=\frac{1-(-1)}{n}=\frac{2}{n} \\ \mathrm{Δ}A=\mathrm{Δ}x×\mathrm{Δ}y=\frac{4}{m}×\frac{2}{n}=\frac{8}{mn} \\ \underset{j=1}{\overset{m}{∑}}\underset{k=1}{\overset{n}{∑}}(-2+j\mathrm{Δ}x)(-1+k\mathrm{Δ}y{)}^3\mathrm{Δ}A= \\ =\mathrm{Δ}A\left(\mathrm{Δ}{y}^3\frac{n^2(n+1{)}^2}{4}-\mathrm{Δ}{y}^2\frac{n(n+1)(2n+1)}{2}+\mathrm{Δ}y\frac{3n(n+1)}{2}-n\right)\left(-2m+\mathrm{Δ}x\frac{m(m+1)}{2}\right) \\ =\left(\frac{8}{mn}\right)\left(\mathrm{Δ}{y}^3\frac{n^2(n+1{)}^2}{4}-\mathrm{Δ}{y}^2\frac{n(n+1)(2n+1)}{2}+\mathrm{Δ}y\frac{3n(n+1)}{2}-n\right)\left(-2m+\mathrm{Δ}x\frac{m(m+1)}{2}\right) \\ =8\left({\left(\frac{2}{n}\right)}^3\frac{n^2(n+1{)}^2}{4}\frac{1}{n}-{\left(\frac{2}{n}\right)}^2\frac{n(n+1)(2n+1)}{2}\frac{1}{n}+\left(\frac{2}{n}\right)\frac{3n(n+1)}{2}\frac{1}{n}-n\frac{1}{n}\right)×\left(-2m\frac{1}{m}+\frac{4}{m}\frac{m(m+1)}{2}\frac{1}{m}\right) \\ =8\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)×\left(-2+\frac{2(m+1)}{m}\right) \end{gathered}$$

Now the double integration is

$$\begin{gathered} {âˆ\neg }_{R}f(x,y)dA=\underset{\mathrm{Δ}→0}{\lim }\underset{j=1}{\overset{m}{∑}}\underset{k=1}{\overset{n}{∑}}(-2+j\mathrm{Δ}x)(-1+k\mathrm{Δ}y{)}^3\mathrm{Δ}A \\ =\underset{\mathrm{Δ}→0}{\lim }8\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)×\left(-2+\frac{2(m+1)}{m}\right) \\ =8\underset{m→∞}{\lim }\left(\underset{n→∞}{\lim }\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)×\left(-2+\frac{2(m+1)}{m}\right)\right) \\ =8\underset{n→∞}{\lim }\left(\underset{m→∞}{\lim }\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)×\left(-2+\frac{2(m+1)}{m}\right)\right) \\ =8\underset{n→∞}{\lim }\left(\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)\underset{m→∞}{\lim }\left(-2+\frac{2(m+1)}{m}\right)\right) \\ =8\underset{n→∞}{\lim }\left(\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)\left(\underset{m→∞}{\lim }-2+\underset{m→∞}{\lim }\frac{2(m+1)}{m}\right)\right) \\ =8\underset{n→∞}{\lim }\left(\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)(-2+2)\right) \\ =8\underset{n→∞}{\lim }\left(\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)\left(0\right)\right) \\ =8\underset{n→∞}{\lim }(0) \\ =0 \end{gathered}$$