Question

Find the value of\(\iint_S zdS\)

Short answer

The value of \(\iint_S zdS\) is\(\frac{{13}}{{12}}\sqrt 2 \).

Step-by-step solution

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

Find \(\left| {{{\rm{r}}_y} \times {{\rm{r}}_z}} \right|\)

Formula used;

\(\iint_S f(x,y,z)dS = \iint_D f(r(u,v))\mid {r_u} \times {r_v}\mid dA\) …………. (1)

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

\({{\rm{r}}_z} = \frac{{\partial x}}{{\partial z}}{\rm{i}} + \frac{{\partial y}}{{\partial z}}{\rm{j}} + \frac{{\partial z}}{{\partial z}}{\rm{k}}\quad \) …………. (3)

Consider the general form of \({\rm{r}}(y,z)\)is \({\rm{r}}(y,z) = x{\rm{i}} + y{\rm{j}} + z{\rm{k}}\).

Substitute \(y + 2{z^2}\) for \(x\) in \({\rm{r}}(y,z) = x{\rm{i}} + y{\rm{j}} + z{\rm{k}}\);

\({\rm{r}}(y,z) = \left( {y + 2{z^2}} \right){\rm{i}} + y{\rm{j}} + z{\rm{k}}\)

Obtain the value of\({{\rm{r}}_{\rm{y}}}\);

Substitute \(y + 2{z^2}\) for \(x\) in equation (2);

\(\begin{array}{l}{{\rm{r}}_y} = \frac{\partial }{{\partial y}}\left( {y + 2{z^2}} \right){\rm{i}} + \frac{{\partial y}}{{\partial y}}{\rm{j}} + \frac{{\partial z}}{{\partial y}}{\rm{k}}\\{{\rm{r}}_y} = {\rm{i}} + {\rm{j}} + (0){\rm{k}}\\{{\rm{r}}_y} = {\rm{i}} + {\rm{j}}\end{array}\)

Obtain the value of \({{\rm{r}}_z}\);

Substitute \(y + 2{z^2}\) for \(x\) in equation (3);

\(\begin{array}{l}{{\rm{r}}_z} = \frac{\partial }{{\partial z}}\left( {y + 2{z^2}} \right){\rm{i}} + \frac{{\partial y}}{{\partial z}}{\rm{j}} + \frac{{\partial z}}{{\partial z}}{\rm{k}}\\{{\rm{r}}_z} = (4z){\rm{i}} + (0){\rm{j}} + (1){\rm{k}}\\{{\rm{r}}_z} = (4z){\rm{i}} + {\rm{k}}\end{array}\)

Obtain the value of \({{\rm{r}}_y} \times {{\rm{r}}_z}\);

\(\begin{array}{l}{{\rm{r}}_{\rm{y}}} \times {{\rm{r}}_z} = ({\rm{i}} + {\rm{j}}) \times ((4z){\bf{i}} + {\rm{k}})\\{{\rm{r}}_{\rm{y}}} \times {{\rm{r}}_z} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\1&1&0\\{4z}&0&1\end{array}} \right|\\{{\rm{r}}_{\rm{y}}} \times {{\rm{r}}_z} = (1 - 0){\rm{i}} - (1 - 0){\rm{j}} + (0 - 4z){\rm{k}}\\{{\rm{r}}_{\rm{y}}} \times {{\rm{r}}_z} = {\rm{i}} - {\rm{j}} - 4z{\rm{k}}\end{array}\)

Obtain the value of \(\left| {{{\bf{r}}_y} \times {{\rm{r}}_z}} \right|\);

\(\begin{array}{l}\left| {{{\rm{r}}_y} \times {{\rm{r}}_z}} \right| = |{\rm{i}} - {\rm{j}} - 4z{\rm{k}}|\\\left| {{{\rm{r}}_y} \times {{\rm{r}}_z}} \right| = \sqrt {{{(1)}^2} + {{( - 1)}^2} + {{( - 4z)}^2}} \\\left| {{{\rm{r}}_y} \times {{\rm{r}}_z}} \right| = \sqrt {1 + 1 + 16{z^2}} \\\left| {{{\rm{r}}_y} \times {{\rm{r}}_z}} \right| = \sqrt {2 + 16{z^2}} \end{array}\)

Obtain the value of 

Modify equation (1) as follows;

(\iint_S zdS = \iint_D z\left( {\left| {{r_y} \times {r_z}} \right|} \right)dA\)

Apply limits and substitute \(\sqrt {2 + 16{z^2}} \) for\(\left| {{r_y} \times {r_z}} \right|\).

\(\begin{aligned}\iint_S zdS &= \int_0^1 {\int_0^1 z } \left( {\sqrt {2 + 16{z^2}} } \right)dzdy \hfill \\\iint_S zdS &= \int_0^1 z \left( {\sqrt {2 + 16{z^2}} } \right)[y]_0^1dz \hfill \\\iint_S zdS &= \int_0^1 z \left( {\sqrt {2 + 16{z^2}} } \right)(1 - 0)dz \hfill \\\iint_S zdS &= \int_0^1 z \left( {\sqrt {2 + 16{z^2}} } \right)dz \hfill \\\end{aligned} \)

Simplify the equation by substitution method;

Let \(u = 2 + 16{z^2}\) then, \(du = 32zdz\).

The limit can be changed as follows;

If \(z = 1\) then, \(u = 18\).

If \(z = 0\) then, \(u = 2\).

\(\begin{aligned}\iint_S zdS &= \frac{1}{{32}}\int_2 3 2z\left( {\sqrt {2 + 16{z^2}} } \right)dz \hfill \\\iint_S zdS &= \frac{1}{{32}}\int_2^{18} {\sqrt u } du \hfill \\\iint_S zdS &= \frac{1}{{32}}\left[ {\frac{2}{3}{{(u)}^{\frac{3}{2}}}} \right]_2^{18} \hfill \\\iint_S zdS &= \frac{1}{{48}}\left[ {{{(18)}^{\frac{3}{2}}} - {{(2)}^{\frac{3}{2}}}} \right] \hfill \\\end{aligned} \)

Modify the equation;

\(\begin{aligned}\iint_S zdS &= \frac{1}{{48}}\left[ {{{(18)}^{\frac{3}{2}}} - {{(2)}^{\frac{3}{2}}}} \right] \hfill \\\iint_S zdS &= \frac{{52}}{{48}}\sqrt 2 \hfill \\\iint_S zdS &= \frac{{13}}{{12}}\sqrt 2 \hfill \\\end{aligned} \)

Thus, the value of is\(\frac{{13}}{{12}}\sqrt 2 \).