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Essential Calculus: Early Transcendentals
Found in: Page 817
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 \(\iint_S ydS\)

The value of \(\iint_S ydS\) is\(\frac{2}{3}(2\sqrt 2 - 1)\).

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 \(\iint_S f(x,y,z)dS = \mathop {\lim }\limits_{\max \Delta {u_i},\Delta {v_j} \to 0} \sum\limits_{i = 1}^m {\sum\limits_{j = 1}^n f } \left( {P_{ij}^*} \right)\Delta {S_{ij}}\) where, \(\Delta {S_{ij}} \approx \left| {{{\bf{r}}_u} \times {{\bf{r}}_v}} \right|\Delta {u_i}\Delta {v_j}\).”

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

As given data;

\(r(u,v) = \langle u\cos v,u\sin v,v\rangle ,0 \le u \le 1\) and \(0 \le v \le \pi .\)

Formula used;

…… (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)

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

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

Substitute \(u\cos v\) for \(x,u\sin v\) for \(y\) and \(v\) for \(z\) in equation (2);

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

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

Substitute \(u\cos v\) for \(x,u\sin v\) for \(y\) and \(v\) for \(z\) in equation (3);

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

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

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

\(\begin{array}{l}\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = |\sin v{\rm{i}} - \cos v{\rm{j}} + u{\rm{k}}|\\\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = \sqrt {{{(\sin v)}^2} + {{\left( { - {{\cos }^2}v} \right)}^2} + {{(u)}^2}} \\\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = \sqrt {{{\sin }^2}v + {{\cos }^2}v + {u^2}} \\\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = \sqrt {\left( {{{\sin }^2}v + {{\cos }^2}v} \right) + {u^2}} \end{array}\)

Simplify the equation;

\(\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = \sqrt {1 + {u^2}} \)

Step 4: Find 

Modify equation (1) as follows;

\(\iint_S ydS = \iint_D y\left( {\left| {{r_u} \times {r_v}} \right|} \right)dA\)

Apply limits and substitute \(u\sin v\) for \(y\) and \(\sqrt {1 + {u^2}} \) for \(\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right|\);

\(\begin{aligned}\iint_S ydS &= \int_0^1 {\int_0^\pi {(u\sin v)} } \left( {\sqrt {1 + {u^2}} } \right)dudv\hfill \\\iint_S ydS &= \int_0^1 u \sqrt {1 + {u^2}} du\int_0^\pi {\sin } vdv \hfill \\\end{aligned} \)

…… (4)

Apply substitution method;

\(\begin{aligned}t &= 1 + {u^2}\\dt &= 0 + 2udu\\dt &= 2udu\end{aligned}\)

Find new limits;

\(\begin{array}{l}{t_{{\rm{upper }}}} = 1 + {(1)^2}\\{t_{{\rm{upper }}}} = 1 + 1\\{t_{{\rm{upper }}}} = 2\end{array}\)

For lower limit,

\(\begin{array}{l}{t_{{\rm{lower }}}} = 1 + {(0)^2}\\{t_{{\rm{lower }}}} = 1 + 0\\{t_{{\rm{lower }}}} = 1\end{array}\)

Apply new limits and substitute \(t\) for \(1 + {u^2},2udu\) for \(dt\);

\(\begin{aligned}\iint_S ydS &= \int_1^2 {\frac{1}{2}} \sqrt t dt\int_0^\pi {\sin } vdv \hfill \\\iint_S ydS &= \frac{1}{2}\int_1^2 {\left( {{t^{\frac{1}{2}}}} \right)} dt\int_0^\pi {\sin } vdv \hfill \\\iint_S ydS &= \frac{1}{2}\left[ {\frac{{{t^{\frac{3}{2}}}}}{{\left( {\frac{3}{2}} \right)}}} \right]_1^2[ - \cos v]_0^\pi \hfill \\\iint_S ydS &= \frac{1}{2}\left( {\frac{2}{3}} \right)\left[ {{t^{\frac{3}{2}}}} \right]_1^2[ - \cos v]_0^\pi \hfill \\\end{aligned} \)

Simplify the equation;

\(\begin{aligned}\iint_S ydS &= \frac{1}{2}\left( {\frac{2}{3}} \right)\left[ {{{(2)}^{\frac{3}{2}}} - {{(1)}^{\frac{3}{2}}}} \right][ - \cos \pi + \cos 0] \hfill \\\iint_S ydS &= \frac{1}{3}\left[ {{{(2)}^{\frac{3}{2}}} - 1} \right][ - ( - 1) + (1)] \hfill \\\iint_S ydS &= \frac{1}{3}\left[ {{{(2)}^{\frac{3}{2}}} - 1} \right](2) \hfill \\\iint_S ydS &= \frac{2}{3}(2\sqrt 2 - 1) \hfill \\\end{aligned} \)

Thus, the value of \(\iint_S ydS\) is\(\frac{2}{3}(2\sqrt 2 - 1)\).

Most popular questions for Math Textbooks

Evaluate the line integral, where C is the given curve.

\(\) \(\int_{\rm{C}} {\rm{x}} {{\rm{y}}^{\rm{4}}}{\rm{ds}}\), C is the right half of the circle \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 16}}\).

Determine whether or not\(F\)is a conservative vectorfield. If it is, find a function\({\rm{f}}\)such that\({\rm{F = }}\nabla {\rm{f}}\).

\({\rm{F}}\left( {{\rm{x,y}}} \right){\rm{ = }}\left( {{\rm{3}}{{\rm{x}}^{\rm{2}}}{\rm{ - 2}}{{\rm{y}}^{\rm{2}}}} \right){\rm{i + }}\left( {{\rm{4xy + 3}}} \right){\rm{j}}\)

A particle moves in a velocity field \(V(x,y) = \left\langle {{x^2},x + {y^2}} \right\rangle \). If it is at position \((2,1)\)at time \(t = 3\), estimate its location at time \(t = 3.01\).

Use the Divergence Theorem to evaluate , where \({\bf{F}}(x,y,z) = {z^2}x{\bf{i}} + \left( {\frac{1}{3}{y^3} + \tan z} \right){\bf{j}} + \left( {{x^2}z + {y^2}} \right){\bf{k}}\)and \(S\) is the top half of the sphere \({x^2} + {y^2} + {z^2} = 1\) . (Hint: Note that \(S\) is not a closed surface. First compute integrals over \({S_1}\) and \({S_2}\), where \({S_1}\) is the disk , oriented downward, and \({S_2} = S \cup {S_1}\).)

To find: The area of surface with the vector equation

\(z = 1 + 2x + 3y + 4{y^2}\)
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