Question

To determine the flux of the given vector field \(F\) across \(S\).

Short answer

The flux of the given vector field \(F\) across \(S\) is \(0\).

Step-by-step solution

Concept of Flux of vector field

If\({\rm{F}}\)is a continuous vector field defined on an oriented surface\({\rm{S}}\)with unit normal vector\({\rm{n}}\), then the surface integral of Fover\({\rm{S}}\)is . This integral is also called the flux of\({\rm{F}}\)across\({\rm{S}}\)

The normal vector\({\rm{n}}\)to the tangent plane with tangent vectors\({{\rm{r}}_u}\)and\({{\rm{r}}_v}\)is\({\rm{n}} = {{\rm{r}}_u} \times {{\rm{r}}_v}\), where\({{\rm{r}}_u} = \frac{{\partial x}}{{\partial u}}\left( {{u_0},{v_0}} \right){\rm{i}} + \frac{{\partial y}}{{\partial u}}\left( {{u_0},{v_0}} \right){\rm{j}} + \frac{{\partial z}}{{\partial u}}\left( {{u_0},{v_0}} \right){\rm{k}}\)and\({{\rm{r}}_v} = \frac{{\partial x}}{{\partial v}}\left( {{u_0},{v_0}} \right){\rm{i}} + \frac{{\partial y}}{{\partial v}}\left( {{u_0},{v_0}} \right){\rm{j}} + \frac{{\partial z}}{{\partial v}}\left( {{u_0},{v_0}} \right){\rm{k}}\)

Calculation of the tangent vector\({{\rm{r}}_u}\)

Let \(x = u\cos t,y = {u^2},z = u\sin t\) and the parameter domain \(0 \le u \le 1,0 \le t \le 2\pi \)

Hence, \({\rm{F}}(x,y,z) = \left\langle {0,{u^2}, - u\sin t} \right\rangle \).

Substitute \(x = u\cos t,y = {u^2},z = u\sin t,v = t\) in \({{\rm{r}}_u} = \frac{{\partial x}}{{\partial u}}\left( {{u_0},{t_0}} \right){\rm{i}} + \frac{{\partial y}}{{\partial u}}\left( {{u_0},{t_0}} \right){\rm{j}} + \frac{{\partial z}}{{\partial u}}\left( {{u_0},{t_0}} \right){\rm{k}}\)

obtain the tangent vectors as follows:

\(\begin{array}{l}{{\rm{r}}_u} = \frac{{\partial x}}{{\partial u}}\left( {{u_0},{t_0}} \right){\rm{i}} + \frac{{\partial y}}{{\partial u}}\left( {{u_0},{t_0}} \right){\rm{j}} + \frac{{\partial z}}{{\partial u}}\left( {{u_0},{t_0}} \right){\rm{k}}\\{{\rm{r}}_u} = \frac{{\partial x}}{{\partial u}}(u\cos t){\rm{i}} + \frac{{\partial y}}{{\partial u}}\left( {{u^2}} \right){\rm{j}} + \frac{{\partial z}}{{\partial u}}(u\sin t){\rm{k}}\\{{\rm{r}}_u} = (\cos t){\rm{i}} + (2u){\rm{j}} + (\sin t){\rm{k}}\end{array}\)

Thus, \({{\rm{r}}_u} = \langle \cos t,2u,\sin t\rangle \).

Calculation of the tangent vector\({{\rm{r}}_t}\)

Substitute \(x = u\cos t,y = {u^2},z = u\sin t,v = t\) in \({{\rm{r}}_t} = \frac{{\partial x}}{{\partial t}}\left( {{u_0},{t_0}} \right){\rm{i}} + \frac{{\partial y}}{{\partial t}}\left( {{u_0},{t_0}} \right){\rm{j}} + \frac{{\partial z}}{{\partial t}}\left( {{u_0},{t_0}} \right){\rm{k}}\) obtain the tangent vectors as follows:

\(\begin{array}{l}{{\rm{r}}_t} = \frac{{\partial x}}{{\partial t}}\left( {{u_0},{t_0}} \right){\rm{i}} + \frac{{\partial y}}{{\partial t}}\left( {{u_0},{t_0}} \right){\rm{j}} + \frac{{\partial z}}{{\partial t}}\left( {{u_0},{t_0}} \right){\rm{k}}\\{{\rm{r}}_t} = \frac{{\partial x}}{{\partial t}}(u\cos t){\rm{i}} + \frac{{\partial y}}{{\partial t}}\left( {{u^2}} \right){\rm{j}} + \frac{{\partial z}}{{\partial t}}(u\sin t){\rm{k}}\\{{\rm{r}}_t} = ( - u\sin t){\rm{i}} + (0){\rm{j}} + (u\cos t){\rm{k}}\\{{\rm{r}}_t} = \langle \cos t,2u,\sin t\rangle \end{array}\)

Thus, \({{\rm{r}}_t} = \langle \cos t,2u,\sin t\rangle \).

Calculation of the normal vector\(n\)

Substitute \({{\rm{r}}_u} = \langle \cos t,2u,\sin t\rangle ,{{\rm{r}}_t} = \langle \cos t,2u,\sin t\rangle \) in \({r_u} \times {{\rm{r}}_v}\)obtain the normal vector to the tangent plane as follows:

\(\begin{array}{l}{\rm{n}} = {{\rm{r}}_u} \times {{\rm{r}}_v}\\{\rm{n}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{\cos t}&{2u}&{\sin t}\\{ - u\sin t}&0&{u\cos t}\end{array}} \right|\\{\rm{n}} = \left( {2{u^2}\cos t} \right){\rm{i}} - \left( {u{{\cos }^2}t + u{{\sin }^2}t} \right){\rm{j}} + \left( {2{u^2}\sin t} \right){\rm{k}}\\{\rm{n}} = \left( {2{u^2}\cos t} \right){\rm{i}} - \left( {u\left( {{{\cos }^2}t + {{\sin }^2}t} \right)} \right){\rm{j}} + \left( {2{u^2}\sin t} \right){\rm{k}}\\{\rm{n}} = \left( {2{u^2}\cos t} \right){\rm{i}} - u{\rm{j}} + \left( {2{u^2}\sin t} \right){\rm{k}}\end{array}\)

Thus, \({\rm{n}} = \left\langle {2{u^2}\cos t, - u,2{u^2}\sin t} \right\rangle \)

Calculation of the integral

Substitute \({\rm{n}} = \left\langle {2{u^2}\cos t, - u,2{u^2}\sin t} \right\rangle ,{\rm{F}}(x,y,z) = \left\langle {0,{u^2}, - u\sin t} \right\rangle \) in obtain the surface integral as shown below:

The above equation can be further simplified as shown below:

Thus, the flux of the given vector field \(F\) across \(S\) is \(0\).