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

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, \(\Delta {S_{ij}} \approx \left| {{{\bf{r}}_u} \times {{\bf{r}}_v}} \right|\Delta {u_i}\Delta {v_j}\).”

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 \({{\rm{r}}_u} \times {{\rm{r}}_v}\);

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

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

Simplify the equation;

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

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 \(\sqrt 2 \pi \).