Question

A contour map is shown for a function \(f\) on the square \(R = \left( {0,4} \right) \times \left( {0,4} \right)\) . Use the Midpoint Rule with \(m = n = 2\) to estimate the value of \(\int {\int {_R} } f(x,y)dA.\)

Short answer

Finding estimated value of \(\int{\int{{}_{\text{R}}}}\text{+}\left( \text{x,y} \right)\text{dA}\)by using midpoint ruler with\(\text{m=n=2}\).

And a contour map is shown for a function + on the surface\(\text{R=}\left( \text{0,4} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{0,4} \right)\).

Step-by-step solution

Step-1: - Finding partitions:

Consider a contour map for a function + on the square\({\rm{R = }}\left( {{\rm{0,4}}} \right){\rm{ \times }}\left( {{\rm{0,4}}} \right)\). To estimate the value of integral\(\int {\int {{}_{\rm{R}}} } {\rm{ + }}\left( {{\rm{x,y}}} \right){\rm{dA}}\), use the midpoint ruler with\({\rm{m = n = 2}}\).

To approximate the integral, use a double Riemann sum.

\(\int {\int {{}_{\rm{R}}} } {\rm{ + }}\left( {{\rm{x,y}}} \right){\rm{dA = }}\sum\limits_{{\rm{i = j}}}^{\rm{m}} {\sum\limits_{{\rm{j = i}}}^{\rm{n}} {\rm{ + }} \left( {{\rm{xij,yij}}} \right){\rm{\Delta Aij}}} \)where m represents the number of partitions along the y integral\(\left( {{\rm{c,d}}} \right)\).

\(\begin{array}{l}{\rm{c}} = {{\rm{y}}_{\rm{0}}}\angle {{\rm{y}}_{\rm{1}}}\angle ....\angle {{\rm{y}}_{{\rm{i - j}}}}\angle {{\rm{y}}_{\rm{i}}}\angle \therefore .\angle {{\rm{y}}_{\rm{n}}} = {\rm{d}}\\{\rm{and}}\,\Delta {{\rm{A}}_{{\rm{ij}}}} = \Delta {\rm{x}}{}_{{\rm{ij}}}\Delta {{\rm{y}}_{{\rm{ij}}}}\end{array}\)

Step-2:- Writing integral as the double Riemann sum:

In this particular problem the x interval is\(\left( {{\rm{0,4}}} \right)\)and

\(\begin{aligned}{l}{\rm{\Delta }}{{\rm{x}}_{{\rm{ij}}}}{\rm{ = }}\frac{{{\rm{4 - 0}}}}{{\rm{2}}}{\rm{ = }}\frac{{\rm{4}}}{{\rm{2}}}\\ \Rightarrow {\rm{\Delta }}{{\rm{x}}_{{\rm{ij}}}}{\rm{ = 2}}\end{aligned}\)

Similarly, say that the y interval is\(\left( {{\rm{0,4}}} \right)\)and

\(\begin{aligned}{l}{\rm{\Delta }}{{\rm{x}}_{{\rm{ij}}}}{\rm{ = }}\frac{{{\rm{4 - 0}}}}{{\rm{2}}}{\rm{ = }}\frac{{\rm{4}}}{{\rm{2}}}\\ \Rightarrow {\rm{\Delta }}{{\rm{x}}_{{\rm{ij}}}}{\rm{ = 2}}\end{aligned}\)

Additionally, \(\begin{aligned}{l}\Delta {{\rm{A}}_{{\rm{ij}}}} = \,\Delta {{\rm{x}}_{{\rm{ij}}}}\Delta {{\rm{y}}_{{\rm{ij}}}}\\ & = \left( 2 \right)\left( 2 \right)\\ & = 4\\ & \Rightarrow \Delta {{\rm{A}}_{{\rm{ij}}}} = 4\end{aligned}\)

This allows us to write the integral as the double Riemann sum.

\(\begin{aligned} \int {\int {{}_R} } + \left( {x,y} \right)dA &= \sum\limits_{i = 1}^2 {\sum\limits_{j = 1}^2 + \left( {xij,yij} \right)4} \\ &= 4\sum\limits_{i = 1}^2 {\sum\limits_{j = 1}^2 + \left( {xij,yij} \right)} \\ &= 4\sum\limits_{i = 1}^2 {\sum\limits_{j = 1}^2 + \left( {xij,yij} \right)} \end{aligned}\)

Choose the same points \( + \left( {{\rm{xij,yij}}} \right)\)

In any manner we wish

Step-3:- Consider sample points:

In this particular problem to apply the midpoint rule which yields

\[\begin{array}{l}{{\rm{x}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{x}}_{\rm{2}}}{\rm{ = 3}}\\{{\rm{y}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{y}}_{\rm{2}}}{\rm{ = 3}}\end{array}\]

Taking the sample points from the given figure as follows

\[\begin{array}{l}{\rm{ + }}\left( {{\rm{1,1}}} \right){\rm{ = 28,}}\\{\rm{ + }}\left( {{\rm{1,3}}} \right){\rm{ = 4,}}\\{\rm{ + }}\left( {{\rm{3,1}}} \right){\rm{ = 15,}}\\{\rm{ + }}\left( {{\rm{3,3}}} \right){\rm{ = 15}}{\rm{.}}\end{array}\]

Step-4:- Finding estimated value of integral:

Then the integral becomes

\(\begin{aligned}& \int{\int{{}_{\text{R}}}}\text{+}\left( \text{x,y} \right)\text{dA=4}\sum\limits_{\text{i=1}}^{\text{2}}{\sum\limits_{\text{j=1}}^{\text{2}}{\text{+}}\left( \text{xij,yij} \right)} \\& \text{=}\left( \text{f}\left( \text{1,1} \right)\text{+f}\left( \text{3,1} \right)\text{+f}\left( \text{1,3} \right)\text{+}\left( \text{3,3} \right) \right)\text{ }\!\!\times\!\!\text{ }\!\!\Delta\!\!\text{ A} \\& \text{=4}\left( \text{28+4+15+15} \right) \\ & \text{=4}\left( \text{62} \right) \\ & \Rightarrow \iint\limits_{\text{D}}{\text{f}\left( \text{x,y} \right)}\text{dA=248} \\\end{aligned}\)

Hence, the required estimated value of the integral

.