Q. 31 - Calculus

Question

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

${\iint }_{R}x{y}^{3}dA$

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

Step-by-Step Solution

Answer

The value of integration is zero.

Step 1. Given information

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

Step 2. Evaluating integral

Use identity 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

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

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

working on the Riemann sum gives,

$$\begin{gathered} \sum _{j=1}^m\sum _{k=1}^n(-2+j\Delta x)(-1+k\Delta y{)}^3\Delta A=\sum _{j=1}^m(-2+j\Delta x)\Delta A\left(\sum _{k=1}^n(-1+k\Delta y{)}^3\right) \\ =\sum _{j=1}^m(-2+j\Delta x\left)\Delta A\left(\sum _{i=1}^n(k\Delta y-1{)}^3\right) \\ \sum _{k=1}^n\right(k\Delta y-1{)}^3=\sum _{i=1}^n\left((k\Delta y{)}^3-3(k\Delta y{)}^{2}1+3k\Delta y(1{)}^2-\left(1^3\right)\right) \\ =\sum _{k=1}^n\left(k^3\Delta y^3-3k^2\Delta y^2+3k\Delta y-1\right) \\ =\sum _{k=1}^n{k}^3\Delta y^3-\sum _{k=1}^{n}3k^2\Delta y^2+\sum _{k=1}^{n}3k\Delta y-\sum _{k=1}^n\left(1\right) \\ =\Delta y^3\sum _{k=1}^n{k}^3-3\Delta y^2\sum _{k=1}^n{k}^2+3\Delta y\sum _{k=1}^{n}k-\sum _{k=1}^n\left(1\right) \\ =\Delta y^3\frac{n^2(n+1{)}^2}{4}-3\Delta y^2\frac{n(n+1)(2n+1)}{6}+3\Delta y\frac{n(n+1)}{2}-n \\ =\Delta y^3\frac{n^2(n+1{)}^2}{4}-\Delta y^2\frac{n(n+1)(2n+1)}{2}+\Delta y\frac{3n(n+1)}{2}-n \end{gathered}$$

Hence,

$$\begin{gathered} \sum _{j=1}^m\sum _{k=1}^n(-2+j\Delta x)(-1+k\Delta y{)}^3\Delta A= \\ =\sum _{j=1}^m(-2+j\Delta x)\Delta A\left(\Delta y^3\frac{n^2(n+1{)}^2}{4}-\Delta y^2\frac{n(n+1)(2n+1)}{2}+\Delta y\frac{3n(n+1)}{2}-n\right) \\ =\Delta A\left(\Delta y^3\frac{n^2(n+1{)}^2}{4}-\Delta y^2\frac{n(n+1)(2n+1)}{2}+\Delta y\frac{3n(n+1)}{2}-n\right)\sum _{j=1}^m(-2+j\Delta x) \\ =\Delta A\left(\Delta y^3\frac{n^2(n+1{)}^2}{4}-\Delta y^2\frac{n(n+1)(2n+1)}{2}+\Delta y\frac{3n(n+1)}{2}-n\right)\left(\sum _{j=1}^m(-2)+\sum _{j=1}^m\left(j\Delta x\right)\right) \\ =\Delta A\left(\Delta y^3\frac{n^2(n+1{)}^2}{4}-\Delta y^2\frac{n(n+1)(2n+1)}{2}+\Delta y\frac{3n(n+1)}{2}-n\right)\left(-2m+\Delta r\frac{m(m+1)}{2}\right) \end{gathered}$$

Recall,

$$\begin{gathered} \Delta x=\frac{b-a}{m}=\frac{2-(-2)}{m}=\frac{4}{m} \\ \Delta y=\frac{d-c}{n}=\frac{1-(-1)}{n}=\frac{2}{n} \\ \Delta A=\Delta x\times \Delta y=\frac{4}{m}\times \frac{2}{n}=\frac{8}{mn} \\ \sum _{j=1}^m\sum _{k=1}^n(-2+j\Delta x)(-1+k\Delta y{)}^3\Delta A= \\ =\Delta A\left(\Delta y^3\frac{n^2(n+1{)}^2}{4}-\Delta y^2\frac{n(n+1)(2n+1)}{2}+\Delta y\frac{3n(n+1)}{2}-n\right)\left(-2m+\Delta x\frac{m(m+1)}{2}\right) \\ =\left(\frac{8}{mn}\right)\left(\Delta y^3\frac{n^2(n+1{)}^2}{4}-\Delta y^2\frac{n(n+1)(2n+1)}{2}+\Delta y\frac{3n(n+1)}{2}-n\right)\left(-2m+\Delta x\frac{m(m+1)}{2}\right) \\ =8\left({\left(\frac{2}{n}\right)}^3\frac{n^2(n+1{)}^2}{4}\frac{1}{n}-{\left(\frac{2}{n}\right)}^2\frac{n(n+1)(2n+1)}{2}\frac{1}{n}+\left(\frac{2}{n}\right)\frac{3n(n+1)}{2}\frac{1}{n}-n\frac{1}{n}\right)\times \left(-2m\frac{1}{m}+\frac{4}{m}\frac{m(m+1)}{2}\frac{1}{m}\right) \\ =8\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)\times \left(-2+\frac{2(m+1)}{m}\right) \end{gathered}$$

Step 3. Further calculation

Now the double integration is 

$$\begin{gathered} {\iint }_{R}f(x,y)dA=\underset{\Delta \to 0}{\lim }\sum _{j=1}^m\sum _{k=1}^n(-2+j\Delta x)(-1+k\Delta y{)}^3\Delta A \\ =\underset{\Delta \to 0}{\lim }8\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)\times \left(-2+\frac{2(m+1)}{m}\right) \\ =8\underset{m\to \infty }{\lim }\left(\underset{n\to \infty }{\lim }\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)\times \left(-2+\frac{2(m+1)}{m}\right)\right) \\ =8\underset{n\to \infty }{\lim }\left(\underset{m\to \infty }{\lim }\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)\times \left(-2+\frac{2(m+1)}{m}\right)\right) \\ =8\underset{n\to \infty }{\lim }\left(\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)\underset{m\to \infty }{\lim }\left(-2+\frac{2(m+1)}{m}\right)\right) \\ =8\underset{n\to \infty }{\lim }\left(\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)\left(\underset{m\to \infty }{\lim }-2+\underset{m\to \infty }{\lim }\frac{2(m+1)}{m}\right)\right) \\ =8\underset{n\to \infty }{\lim }\left(\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)(-2+2)\right) \\ =8\underset{n\to \infty }{\lim }\left(\left(\frac{2(n+1{)}^2}{n^2}-\frac{2(n+1)(2n+1)}{n^2}+\frac{3(n+1)}{n}-1\right)\left(0\right)\right) \\ =8\underset{n\to \infty }{\lim }(0) \\ =0 \end{gathered}$$