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Found in: Page 817

### Essential Calculus: Early Transcendentals

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

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# 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 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$$\frac{2}{3}(2\sqrt 2 - 1)$$.

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