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Found in: Page 817

Essential Calculus: Early Transcendentals

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

Find the value of $$\iint_S {\left( {{x^2} + {y^2}} \right)}dS$$

The value of $$\iint_S {\left( {{x^2} + {y^2}} \right)}dS$$ is $$\sqrt 2 \pi$$.

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;

$$r(u,v)$$$$= \left\langle {2uv,{u^2} - {v^2},{u^2} + {v^2}} \right\rangle {\rm{ and }}{u^2} + {v^2} \le 1.$$

Formula used;

$$\iint_S f(x,y,z)dS = \iint_D f(r(u,v))\left| {{r_u} \times {r_v}} \right|dA$$ ..…. (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)

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

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

Substitute $$2uv$$ for $$x,{u^2} - {v^2}$$ for $$y$$ and $${u^2} + {v^2}$$ for $$z$$ in equation (2);

$$\begin{array}{l}{{\bf{r}}_u} = \frac{\partial }{{\partial u}}(2uv){\rm{i}} + \frac{\partial }{{\partial u}}\left( {{u^2} - {v^2}} \right){\rm{j}} + \frac{\partial }{{\partial u}}\left( {{u^2} + {v^2}} \right){\rm{k}}\\{{\bf{r}}_u} = (2v){\rm{i}} + (2u - 0){\rm{j}} + (2u + 0){\rm{k}}\\{{\bf{r}}_u} = 2v{\rm{i}} + 2u{\rm{j}} + 2u{\rm{k}} = \langle 2v,2u,2u\rangle \end{array}$$

For find the value of $${r_v}$$;

Substitute$$2uv$$ for $$x,{u^2} - {v^2}$$ for $$y$$ and $${u^2} + {v^2}$$ for $$z$$ in equation (3);

$$\begin{array}{l}{{\bf{r}}_v} = \frac{\partial }{{\partial v}}(2uv){\rm{i}} + \frac{\partial }{{\partial v}}\left( {{u^2} - {v^2}} \right){\rm{j}} + \frac{\partial }{{\partial v}}\left( {{u^2} + {v^2}} \right){\rm{k}}\\{{\bf{r}}_v} = (2u){\rm{i}} + (0 - 2v){\rm{j}} + (0 + 2v){\rm{k}}\\{{\bf{r}}_v} = 2u{\rm{i}} - 2v{\rm{j}} + 2v{\rm{k}}\\{{\bf{r}}_v} = \langle 2u, - 2v,2v\rangle \end{array}$$

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

$$\begin{array}{l}{{\rm{r}}_u} \times {{\rm{r}}_v} = \langle 2v,2u,2u\rangle \times \langle 2u, - 2v,2v\rangle \\{{\rm{r}}_u} \times {{\rm{r}}_v} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{2v}&{2u}&{2u}\\{2u}&{ - 2v}&{2v}\end{array}} \right|\\{{\rm{r}}_u} \times {{\rm{r}}_v} = (4uv + 4uv){\rm{i}} - \left( {4{v^2} - 4{u^2}} \right){\rm{j}} + \left( { - 4{v^2} - 4{u^2}} \right){\rm{k}}\\{{\rm{r}}_u} \times {{\rm{r}}_v} = 8uv{\rm{i}} - \left( {4{v^2} - 4{u^2}} \right){\rm{j}} + \left( { - 4{v^2} - 4{u^2}} \right){\rm{k}}\\{{\rm{r}}_u} \times {{\rm{r}}_v} = \left\langle {8uv,\left( {4{v^2} - 4{u^2}} \right),\left( { - 4{v^2} - 4{u^2}} \right)} \right\rangle \end{array}$$

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

$$\begin{array}{l}\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = \left| {8uv,\left( {4{v^2} - 4{u^2}} \right),\left( { - 4{v^2} - 4{u^2}} \right)} \right|\\\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = \sqrt {{{(8uv)}^2} + {{\left( {4{v^2} - 4{u^2}} \right)}^2} + {{\left( { - 4{v^2} - 4{u^2}} \right)}^2}} \\\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = \sqrt {64{u^2}{v^2} + \left( {16{v^4} - 32{v^2}{u^2} + 16{u^4}} \right) + \left( {16{v^4} + 32{v^2}{u^2} + 16{u^4}} \right)} \\\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right| = \sqrt {64{u^2}{v^2} + 32{v^4} + 32{u^4}} \end{array}$$

Simplify the equation;

\begin{aligned}\left| {{r_u} \times {r_v}} \right| &= \sqrt {32\left( {2{u^2}{v^2} + {v^4} + {u^4}} \right)} \hfill \\\left| {{r_u} \times {r_v}} \right| &= \sqrt {32{{\left( {{u^2} + {v^2}} \right)}^2}} \left\{ {\because {{(a + b)}^2} = {a^2} + {b^2} + 2ab} \right\} \hfill \\\left| {{r_u} \times {r_v}} \right| &= 4\sqrt 2 \left( {{u^2} + {v^2}} \right) \hfill \\\end{aligned}

Step 4: Find

Modify equation (1) as follows;

$$\iint_S {\left( {{x^2} + {y^2}} \right)}dS = \iint_D {\left( {{x^2} + {y^2}} \right)}\left( {\left| {{r_u} \times {r_v}} \right|} \right)dA$$

Substitute $$2uv$$for $$x,{u^2} - {v^2}$$ for $$y$$ and $$4\sqrt 2 \left( {{u^2} + {v^2}} \right)$$ for $$\left| {{{\rm{r}}_u} \times {{\rm{r}}_v}} \right|$$,

\begin{aligned}\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= \iint_D {\left( {{{(2uv)}^2} + {{\left( {{u^2} - {v^2}} \right)}^2}} \right)}\left( {4\sqrt 2 \left( {{u^2} + {v^2}} \right)} \right)dA \hfill \\\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= \iint_D {\left( {4{u^2}{v^2} + \left( {{u^4} - 2{u^2}{v^2} + {v^4}} \right)} \right)}\left( {4\sqrt 2 \left( {{u^2} + {v^2}} \right)} \right)dA \hfill \\\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= 4\sqrt 2 \iint_D {\left( {{u^4} + 2{u^2}{v^2} + {v^4}} \right)}\left( {{u^2} + {v^2}} \right)dA \hfill \\\end{aligned}

$$\iint_S {\left( {{x^2} + {y^2}} \right)}dS = 4\sqrt 2 \iint_D {{{\left( {{u^2} + {v^2}} \right)}^3}}dA$$ …… (4)

Let $$u = r\cos \theta ,v = r\sin \theta$$ and limits for $$r$$ and $$\theta$$ are 0 to 1 and 0 to $$2\pi$$.

Modify the equation (4) as follows;

\begin{aligned}\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= 4\sqrt 2 \int_0^{2\pi } {} {\int_0^1 {\left[ {{{(r\cos \theta )}^2} + {{(r\sin \theta )}^2}} \right]} ^3}rdrd\theta \hfill \\\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= 4\sqrt 2 \int_0^{2\pi } {} {\int_0^1 {\left[ {{r^2}\cos {\theta ^2} + {r^2}\sin {\theta ^2}} \right]} ^3}rdrd\theta \hfill \\\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= 4\sqrt 2 \int_0^{2\pi } {} {\int_0^1 {\left[ {{r^2}(\cos {\theta ^2} + \sin {\theta ^2})} \right]} ^3}rdrd\theta \hfill \\\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= 4\sqrt 2 \int_0^{2\pi } {} {\int_0^1 {\left[ {{r^2}} \right]} ^3}rdrd\theta \left\{ {\because \cos {\theta^2} + \sin {\theta ^2} = 1} \right\} \hfill \\\end{aligned}

Simplify the equation;

\begin{aligned}\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= 4\sqrt 2 \int_0^{2\pi } d \theta \int_0^1 {{r^7}} dr\hfill \\\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= 4\sqrt 2 [\theta ]_0^{2\pi }\left[ {\frac{{{r^8}}}{}} \right]_0^1 \hfill \\\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= 4\sqrt 2 [(2\pi ) - 0]\left[ {\frac{1}{8}\left( {{1^8} - {0^8}} \right)} \right] \hfill \\\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= 4\sqrt 2 (2\pi )\left( {\frac{1}{8}} \right) \hfill \\\iint_S {\left( {{x^2} + {y^2}} \right)}dS &= \sqrt 2 \pi \hfill \\\end{aligned}

Thus, the value of $$\iint_S {\left( {{x^2} + {y^2}} \right)}dS$$ is $$\sqrt 2 \pi$$.