Question

Question: Find the flux of \(\({\bf{F}}(x,y,z) = \sin (xyz){\bf{i}} + {x^2}y{\bf{j}} + {z^2}{e^{x/5}}{\bf{k}}\)\)

across the part of the cylinder \(4{y^2} + {z^2} = 4\) that lies above the \(xy\)-plane and between the planes \(x = - 2\)and \(x = 2\) with upward orientation. Illustrate by using a computer algebra system to draw the cylinder and the vector field on the same screen.

Short answer

The Fluxis\(\frac{{16\pi + 160\sinh (2/5)}}{3}\).

Step-by-step solution

Concept of Surface Integral and Formula Used.

A surface integral is a surface integration generalisation of multiple integrals. It can be thought of as the line integral's double integral equivalent. It is possible to integrate a scalar field or a vector field over a surface.

Definition used:

The value of d S is given by \(dS = \sqrt {1 + {{\left( {\frac{{dz}}{{dx}}} \right)}^2} + {{\left( {\frac{{dz}}{{dy}}} \right)}^2}} dxdy\), that is \(dS = \sqrt {1 + {{\left( {\frac{{dz}}{{dx}}} \right)}^2} + {{\left( {\frac{{dz}}{{\partial y}}} \right)}^2}} dxdy\).

Parameterisation of the cylinder.

The parameterisation of the cylinder is

\(S:{\bf{r}}(u,v) = u{\bf{i}} + \sin v{\bf{j}} + 2\cos v{\bf{k}}\)

Since the surface is part of the cylinder above\(xy\)plane, we have\(z > 0\)

Therefore

\( - \frac{\pi }{2} < v < \frac{\pi }{2}\)

Also the surface is part of the cylinder between the planes\(x = - 2\)and\(x = 2\)

Hence\( - 2 < u < 2\)

Note that

Where\({\bf{n}}\)is the unit normal vector.

Note that\(dS = \left| {{{\bf{r}}_u} \times {{\bf{r}}_v}} \right|dA\)

The Flux through a surface is defined only if the surface is orientable.

Note that if the surface is orientable, there will be two normal unit vectors at every point .

We can write

\(d{\bf{S}} = {\bf{n}}dA = \pm \frac{{\left( {{{\bf{r}}_u} \times {{\bf{r}}_v}} \right)}}{{\left| {{{\bf{r}}_u} \times {{\bf{r}}_v}} \right|}}\left| {{{\bf{r}}_u} \times {{\bf{r}}_v}} \right|dA = \pm \left( {{{\bf{r}}_u} \times {{\bf{r}}_v}} \right)dA\)

For finding the flux we always use\({\bf{n}}\)which correspond to the positive orientation.

In this case the normal vectors must be pointing outwards .

We will first find\({{\bf{r}}_u} \times {{\bf{r}}_v}\)and then we will decide\(d{\bf{S}}\)

\(S:{\bf{r}}(u,v) = u{\bf{i}} + \sin v{\bf{j}} + 2\cos v{\bf{k}}\)

Therefore,

\({{\bf{r}}_u} = {\bf{i}}\)

and

\(\begin{array}{l}{{\bf{r}}_v} = \cos v{\bf{j}} - 2\sin v{\bf{k}}{{\bf{r}}_u} \times {{\bf{r}}_v}\\ = \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\1&0&0\\0&{\cos v}&{ - 2\sin v}\end{array}} \right|{{\bf{r}}_u} \times {{\bf{r}}_v}\\ = {\bf{i}}\left| {\begin{array}{*{20}{c}}0&0\\{\cos v}&{ - 2\sin v}\end{array}} \right| - {\bf{j}}\left| {\begin{array}{*{20}{c}}1&0\\0&{ - 2\sin v}\end{array}} \right| + {\bf{k}}\left| {\begin{array}{*{20}{c}}1&0\\0&{\cos v}\end{array}} \right|{{\bf{r}}_u} \times {{\bf{r}}_v}\\ = 2\sin v{\bf{j}} + \cos v{\bf{k}}\end{array}\)

Calculate the Flux.

Note that \({{\bf{r}}_u} \times {{\bf{r}}_v}\) is pointing outwards

Therefore

\(d{\bf{S}} = \left( {{{\bf{r}}_u} \times {{\bf{r}}_v}} \right)dA = (2\sin v{\bf{j}} + \cos v{\bf{k}})dA\)

\(\begin{array}{l}{\bf{F}}(x,y,z) = \sin (xyz){\bf{i}} + {x^2}y{\bf{j}} + {z^2}{e^{x/5}}{\bf{kF}}({\bf{r}}(u,v))\\ = \sin (2u\sin v\cos v){\bf{i}} + {u^2}\sin v{\bf{j}} + 4{e^{u/5}}{\cos ^2}v{\bf{kF}}({\bf{r}}(u,v))\\ = \sin (u\sin 2v){\bf{i}} + {u^2}\sin v{\bf{j}} + 4{\cos ^2}v{e^{u/5}}{\bf{k}}\end{array}\)

Therefore

Note that \(u \in ( - 2,2)\) and \(v \in ( - \pi /2,\pi /2)\)

\(\begin{array}{l}{{\bf{I}}_1} = \int_{ - \pi /2}^{\pi /2} {{{\sin }^2}} vdv = \frac{1}{2}\int_{ - \pi /2}^{\pi /2} 1 - \cos 2vdv\\ = \frac{1}{2}\left( {v - \frac{{\sin 2v}}{2}} \right)_{ - \pi /2}^{\pi /2}\\ = \frac{\pi }{2}\end{array}\)

\(\begin{array}{l}{{\bf{I}}_2} = \int_{ - \pi /2}^{\pi /2} {{{\cos }^3}} vdv = \int_{ - \pi /2}^{\pi /2} {\left( {{{\cos }^2}v} \right)} \cos vdv\\ = \int_{ - \pi /2}^{\pi /2} {\left( {1 - {{\sin }^2}v} \right)} \cos vdv\end{array}\)

Limits of integration will change from \(\int_ - - \pi /{2^{\pi /2}}\) to \(\mathop \smallint \nolimits_{ - 1}^1 \)

This can be Rewritten as

Thus , the Fluxis\(\frac{{16\pi + 160\sinh (2/5)}}{3}\).