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Essential Calculus: Early Transcendentals
Found in: Page 816
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Find the value of \(\iint_{S}{f}(x,y,z)dS \).

The value of \(\iint_{S}{f}(x,y,z)dS \). is \(11\sqrt {14} \).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Concept of surface integral 

The surface integral is a generalization of multiple integrals that allows for surface integration. The surface integral is sometimes referred to as the double integral. We can integrate across a surface in either the scalar or vector fields for any given surface. The function returns the scalar value in the scalar field and function returns the vector value in the vector field.

“The surface integral of \(f\)over the surface \(S\) as where, \(\iint_S f(x,y,z)dS = \mathop {\lim }\limits_{\max \Delta {u_i},\Delta {v_j} \to 0} \sum\limits_{i = 1}^m {\sum\limits_{j = 1}^n f } \left( {P_{ij}^*} \right)\Delta {S_{ij}}\) where,\(\Delta {S_{ij}} \approx \left| {{r_u} \times {r_v}} \right|\Delta {u_i}\Delta {v_j}\)

Step 2: Find the value of \({{\rm{r}}_u} \times {{\rm{r}}_v}\)

As given data;

\(x = u + v,y = u - v,z = 1 + 2u + v,0 \le u \le 2\) and \(0 \le v \le 1.\)

Use the formula;

\(\iint_{S}{f}(x,y,z)dS=\iint_{D}{f}(r(u,v))\left| {{\text{r}}_{u}}\times {{\text{r}}_{v}} \right|dA\) ..…. (1)

\({{\rm{r}}_u} = \frac{{\partial x}}{{\partial u}}{\rm{i}} + \frac{{\partial y}}{{\partial u}}{\rm{j}} + \frac{{\partial z}}{{\partial u}}{\rm{k}}\) ..…. (2)

\({{\rm{r}}_v} = \frac{{\partial x}}{{\partial v}}{\rm{i}} + \frac{{\partial y}}{{\partial v}}{\rm{j}} + \frac{{\partial z}}{{\partial v}}{\rm{k}}\) …… (3)

For find the value of \({r_u}\);

Substitute \(u + v\) for \(x,u - v\) for \(y\) and \(1 + 2u + v\) for \(z\) in equation (2), then it gives

\(\begin{array}{l}{{\rm{r}}_u} = \frac{\partial }{{\partial u}}(u + v){\rm{i}} + \frac{\partial }{{\partial u}}(u - v){\rm{j}} + \frac{\partial }{{\partial u}}(1 + 2u + v){\rm{k}}\\{{\rm{r}}_u} = (1 + 0){\rm{i}} + (1 - 0){\rm{j}} + (0 + 2 + 0){\rm{k}}\\{{\rm{r}}_u} = {\rm{i}} + {\rm{j}} + 2{\rm{k}}\end{array}\)

For find the value of \({{\rm{r}}_v}\);

Substitute \(u + v\) for \(x,u - v\) for \(y\) and \(1 + 2u + v\) for \(z\) in equation (3);

\(\begin{array}{l}{{\rm{r}}_v} = \frac{\partial }{{\partial v}}(u + v){\rm{i}} + \frac{\partial }{{\partial v}}(u - v){\rm{j}} + \frac{\partial }{{\partial v}}(1 + 2u + v){\rm{k}}\\{{\rm{r}}_v} = (0 + 1){\rm{i}} + (0 - 1){\rm{j}} + (0 + 0 + 1){\rm{k}}\\{{\rm{r}}_v} = {\rm{i}} - {\rm{j}} + {\rm{k}}\end{array}\)

For find the value of \({{\rm{r}}_u} \times {{\rm{r}}_v}\);

\(\begin{array}{l}{{\bf{r}}_u} \times {{\bf{r}}_v} = ({\rm{i}} + {\rm{j}} + 2{\rm{k}}) \times ({\rm{i}} - {\rm{j}} + {\rm{k}})\\{{\bf{r}}_u} \times {{\bf{r}}_v} = \left| {\begin{array}{*{20}{c}}i&j&k\\1&1&2\\1&{ - 1}&1\end{array}} \right|\\{{\bf{r}}_u} \times {{\bf{r}}_v} = (1 + 2){\rm{i}} - (1 - 2){\rm{j}} + ( - 1 - 1){\rm{k}}\\{{\bf{r}}_u} \times {{\bf{r}}_v} = 3{\rm{i}} + {\rm{j}} - 2{\rm{k}}\end{array}\)

Step 3: Find \(\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right|\)

\(\begin{aligned}{l}\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = |3{\rm{i}} + {\rm{j}} - 2{\rm{k}}|\\\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = \sqrt {{{(3)}^2} + {{(1)}^2} + {{( - 2)}^2}} \\\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = \sqrt {9 + 1 + 4} \\\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = \sqrt {14} \end{aligned}\)

Step 4: Find 

Modify equation (1) as follows;

\(\Delta {S_{ij}} \approx \left| {{r_u} \times {r_v}} \right|\Delta {u_i}\Delta {v_j}\)

Apply limits and substitute \(u + v\) for \(x,u - v\) for \(y,1 + 2u+ v\) for \(z\) and \(\sqrt {14} \) for \(\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right|\);

Simplify the equation;

\(\begin{aligned}\iint_S {(x + y + z)}dS &= \int_0^1 {\int_0^2 {(u + v + u - v + 1 + 2u + v)} } (\sqrt {14} )dudv\hfill \\\iint_S {(x + y + z)}dS &= (\sqrt {14} )\int_0^1 {\int_0^2 {(4u + v + 1)} } dudv \hfill \\\iint_S {(x + y + z)}dS &= (\sqrt {14} )\int_0^1 {\left[ {4\left( {\frac{{{u^2}}}{2}} \right) + uv + u} \right]_0^2} dv \hfill \\\iint_S {(x + y + z)}dS &= (\sqrt {14} )\int_0^1 {\left[ {2{u^2} + uv + u} \right]_0^2} dv \hfill \\\end{aligned}\)

Modify the equation;

\(\begin{aligned}\mathop \iint \nolimits_S (x + y + z)dS &= (\sqrt {14} )\smallint _0^1\left[ {2\left( {{2^2} - {0^2}} \right) + v(2 - 0) + (2 - 0)} \right]dv \hfill \\\mathop \iint \nolimits_S (x + y + z)dS &=(\sqrt {14} )\smallint _0^1[2(4) + 2v + 2]dv \hfill \\\mathop \iint \nolimits_S (x+y+z)dS &=(\sqrt {14} )\smallint _0^1[8 + 2v + 2]dv \hfill \\\mathop \iint \nolimits_S (x+y+z)dS &=(\sqrt {14} )\smallint _0^1[2v + 10]dv \hfill \\\end{aligned} \)

Thus, the value of \(\iint_{S}{(x+y+z)}dS \). is\(11\sqrt {14} \).

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