Chapter 16.6, Problem 55E

To determine

(a)

To use:

The midpoint rule for double integral with six squares to estimate the area of the given surface

Solution:

$24.2055$

Explanation:

1) Concept:

Use the formula for surface area and midpoint rule to evaluate it.

2) Formula and rule:

i) The surface area is

$A\left(S\right)={\iint }_D^{\square }\sqrt{1+{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2}dA$

ii) Midpoint rule:

${\iint }_R^{\square }f\left(x, y\right)dA\approx \sum _{i=1}^m\sum _{j=1}^{n}f\left(\stackrel{-}{x_i},\stackrel{-}{y_j}\right)∆A$

Where $\stackrel{-}{x_i}$ is the midpoint of $[x_{i-1}, x_i]$ and $\stackrel{-}{y_i}$ is the midpoint of $[y_{i-1}, y_i]$.

3) Given:

$z=\frac{1}{1+x^2+y^2}, 0\le x\le 6, 0\le y\le 4$

4) Calculation:

Consider the given surface,

$z=\frac{1}{1+x^2+y^2}$

Differentiate $z$ with respect to $x$,

$\frac{\partial z}{\partial x}=\frac{2x}{{\left(1+x^2+y^2\right)}^2}$

Differentiate $z$ with respect to $y$,

$\frac{\partial z}{\partial x}=\frac{2y}{{\left(1+x^2+y^2\right)}^2}$

The area of the surface is given by,

$A\left(S\right)={\iint }_D^{\square }\sqrt{1+{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2}dA$

Substitute the values,

$={\int }_0^6{\int }_0^4\sqrt{1+{\left(\frac{2x}{{\left(1+x^2+y^2\right)}^2}\right)}^2+{\left(\frac{2y}{{\left(1+x^2+y^2\right)}^2}\right)}^2}dydx$

Simplify,

$={\int }_0^6{\int }_0^4\sqrt{1+\frac{4x^2}{{\left(1+x^2+y^2\right)}^4}+\frac{4y^2}{{\left(1+x^2+y^2\right)}^4}}dydx$

$={\int }_0^6{\int }_0^4\sqrt{1+\frac{4x^2+4y^2}{{\left(1+x^2+y^2\right)}^4}}dydx$

By using midpoint rule with

$f\left(x, y\right)=\sqrt{1+\frac{4x^2+4y^2}{{\left(1+x^2+y^2\right)}^4}}, m=3, n=2.$

$\stackrel{-}{x_1}=\frac{0+2}{2}=1$

$\stackrel{-}{x_2}=\frac{2+4}{2}=3$

$\stackrel{-}{x_3}=\frac{4+6}{2}=5$

$\stackrel{-}{y_1}=\frac{0+2}{2}=1$

$\stackrel{-}{y_2}=\frac{2+4}{2}=3$

Area of each sub rectangle is $∆A=4$.

Therefore,

$A\left(S\right)\approx \sum _{i=1}^3\sum _{j=1}^{2}f\left(\stackrel{-}{x_i},\stackrel{-}{y_j}\right)∆A$

Substituting the values,

$=4\left[f\left(\stackrel{-}{x_1},\stackrel{-}{y_1}\right)+f\left(\stackrel{-}{x_1},\stackrel{-}{y_2}\right)+f\left(\stackrel{-}{x_2},\stackrel{-}{y_1}\right)+f\left(\stackrel{-}{x_2},\stackrel{-}{y_2}\right)+f\left(\stackrel{-}{x_3},\stackrel{-}{y_1}\right)+f\left(\stackrel{-}{x_3},\stackrel{-}{y_3}\right)\right]$

$=4\left[f\left(1, 1\right)+f\left(\mathrm{1,3}\right)+f\left(\mathrm{3,1}\right)+f\left(\mathrm{3,3}\right)+f\left(\mathrm{5,1}\right)+f\left(\mathrm{5,3}\right)\right]$

$=4\left[\sqrt{1+\frac{4{\left(1\right)}^2+4{\left(1\right)}^2}{{\left(1+{\left(1\right)}^2+{\left(1\right)}^2\right)}^4}}+\sqrt{1+\frac{4{\left(1\right)}^2+4{\left(3\right)}^2}{{\left(1+{\left(1\right)}^2+{\left(3\right)}^2\right)}^4}}+\sqrt{1+\frac{4{\left(3\right)}^2+4{\left(1\right)}^2}{{\left(1+{\left(3\right)}^2+{\left(1\right)}^2\right)}^4}}+\sqrt{1+\frac{4{\left(3\right)}^2+4{\left(3\right)}^2}{{\left(1+{\left(3\right)}^2+{\left(3\right)}^2\right)}^4}}+\sqrt{1+\frac{4{\left(5\right)}^2+4{\left(1\right)}^2}{{\left(1+{\left(5\right)}^2+{\left(1\right)}^2\right)}^4}}+\sqrt{1+\frac{4{\left(5\right)}^2+4{\left(3\right)}^2}{{\left(1+{\left(5\right)}^2+{\left(3\right)}^2\right)}^4}}\right]$

By using calculator,

$\approx 24.2055$

Conclusion:

The area of the given surface is

$A\left(S\right)\approx 24.2055$

(b)

To use:

A computer algebra system to approximate the surface area in part (a) to four decimal places and compare it to the answer in part (a)

Solution:

$A\left(S\right)\approx 24.2476$, by comparing it with answer to part (a), it is correct up to first decimal place.

Explanation:

1) Concept:

Use computer algebra system to evaluate the surface area and compare it to the answer in part (a).

2) Given:

$z=\frac{1}{1+x^2+y^2}, 0\le x\le 6, 0\le y\le 4$

3) Calculation:

From the part (a),

$A\left(S\right)={\iint }_D^{\square }\sqrt{1+{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2}dA$

$={\int }_0^6{\int }_0^4\sqrt{1+\frac{4x^2+4y^2}{{\left(1+x^2+y^2\right)}^4}}dydx$

By using Mathematica,

Input:

$\mathrm{N}\mathrm{I}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\left[\mathrm{S}\mathrm{q}\mathrm{r}\mathrm{t}\left[1+\left(4*x^2+4*y^2\right)/{\left(1+x^2+y^2\right)}^4\right],\left\{x,\mathrm{0,6}\right\},\left\{y,\mathrm{0,4}\right\}\right]$

Output:

$24.247647165851532$

Therefore, the surface area to four decimal places is

$A\left(S\right)\approx 24.2476$

By comparing it with answer to part (a), it is correct up to first decimal place.

Conclusion:

$A\left(S\right)\approx 24.2476$, by comparing it to the answer in part (a); it is correct up to first decimal place.