Question

Evaluate the surface integral.

\(\iint_{S}{F}\cdot dS\) , where\({\rm{F(x,y,z) = xzi - 2yj + 3xk}}\) and \({\rm{S}}\) is the sphere \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 4}}\) with outward orientation.

Short answer

The surface integral \(\iint_{S}{F}\cdot dS=-\frac{64\pi }{3}\).

Step-by-step solution

Explanation of solution.

Note

\(\iint_{S}{F}\cdot dS=\iint_{D}{F}\cdot ndS\)

The unit normal vector is defined by n.

Note \({\rm{dS = }}\left| {{{\rm{r}}_{\rm{u}}}{\rm{ * }}{{\rm{r}}_{\rm{v}}}} \right|{\rm{dA}}\).

Only if a surface is orientable is the flux through it specified.

It's important to note that if the surface is orientable, each point will have two normal unit vectors.

So

\(dS=ndA=\pm \frac{\left( {{r}_{u}}*{{r}_{v}} \right)}{\left| {{r}_{u}}*{{r}_{v}} \right|}\left| {{r}_{u}}*{{r}_{v}} \right|dA=\pm \left( {{r}_{u}}*{{r}_{v}} \right)dA\).

Step 2:The parameterization for the sphere.

\({\rm{r(\theta }},\phi {\rm{) = 2sin}}\phi {\rm{cos\theta i + 2sin}}\phi {\rm{sin\theta j + 2cos}}\phi {\rm{k}}\)

Were \({\rm{\theta }} \in {\rm{(0,2\pi )}}\) and \({\rm{f}} \in {\rm{(0,\pi )}}\)

\(\begin{aligned}{{\rm{r}}_{\rm{\theta }}}\rm &= - 2 sin\phi {\rm{sin \theta i + 2 sin}}\phi {\rm{cos \theta j}}\\{{\rm{r}}_\phi }\rm &= 2 cos\phi {\rm{cos \theta i + 2 cos}}\phi {\rm{sin\theta j - 2 sin}}\phi {\rm{k}}\end{aligned}\)

\(\begin{aligned}{{\rm{r}}_{\rm{\theta }}}{\rm{ * }}{{\rm{r}}_\phi } &= \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{\rm{ - 2sin}}\phi {\rm{sin\theta }}}&{{\rm{2sin}}\phi {\rm{cos\theta }}}&0\\{{\rm{2cos}}\phi {\rm{cos\theta }}}&{{\rm{2cos}}\phi {\rm{sin\theta }}}&{{\rm{ - 2sin}}\phi }\end{array}} \right|\\{{\rm{r}}_{\rm{\theta }}}{\rm{ * }}{{\rm{r}}_\phi }\rm &= i\left| {\begin{array}{*{20}{c}}{{\rm{2sin}}\phi {\rm{cos\theta }}}&{\rm{0}}\\{{\rm{2cos}}\phi {\rm{sin\theta }}}&{{\rm{ - 2sin}}\phi }\end{array}} \right|{\rm{ - j}}\left| {\begin{array}{*{20}{c}}{{\rm{ - 2sin}}\phi {\rm{sin\theta }}}&{\rm{0}}\\{{\rm{2cos}}\phi {\rm{cos\theta }}}&{{\rm{ - 2sin}}\phi }\end{array}} \right|{\rm{ + k}}\left| {\begin{array}{*{20}{c}}{{\rm{ - 2sin}}\phi {\rm{sin\theta }}}&{{\rm{2sin}}\phi {\rm{cos\theta }}}\\{{\rm{2cos}}\phi {\rm{cos\theta }}}&{{\rm{2cos}}\phi {\rm{sin\theta }}}\end{array}} \right|\\{{\rm{r}}_{\rm{\theta }}}{\rm{ * }}{{\rm{r}}_\phi }\rm &= - 4si{{\rm{n}}^{\rm{2}}}\phi {\rm{cos\theta i - 4si}}{{\rm{n}}^{\rm{2}}}\phi {\rm{sin\theta j - 4sin}}\phi {\rm{cos}}\phi {\rm{k}}\end{aligned}\)

A note \({{\rm{r}}_{\rm{\theta }}}{\rm{ * }}{{\rm{r}}_\phi }\) is pointing towards the origin.

\({\rm{dS = - }}\left( {{{\rm{r}}_{\rm{\theta }}}{\rm{ * }}{{\rm{r}}_\phi }} \right){\rm{dA = }}\left( {{\rm{4si}}{{\rm{n}}^{\rm{2}}}\phi {\rm{cos\theta i + 4si}}{{\rm{n}}^{\rm{2}}}\phi {\rm{sin\theta j + 4sin}}\phi {\rm{cos}}\phi {\rm{k}}} \right){\rm{dA}}\).

\(\begin{aligned}\rm F(x,y,z) &= xzi - 2yj + 3xk\\{\rm{F(r(\theta ,}}\phi \rm )) &= 4sin\phi {\rm{cos}}\phi {\rm{cos\theta i - 4sin}}\phi {\rm{sin\theta j + 6sin}}\phi {\rm{cos\theta k}}{\rm{.}}\end{aligned}\)

Add the red and blue equations.

\(\iint_{S}{F} * dS=\int_{0}^{2\pi }{\int_{0}^{\pi }{1}}6{{\sin }^{3}}\phi \cos \phi {{\cos }^{2}}\theta -16{{\sin }^{3}}\phi {{\sin }^{2}}\theta +24{{\sin }^{2}}\phi \cos \phi \cos \theta d\phi d\theta \)

Use the property

\(\int_{\rm{a}}^{\rm{b}} {\rm{f}} {\rm{(x)dx = }}\int_{\rm{a}}^{\rm{b}} {\rm{f}} {\rm{(a + b - x)dx}}\)

Substitute \({\rm{a + b - x = u}}\)

Note \({\rm{sin(\pi - }}\phi {\rm{) = sin}}\phi {\rm{ }}\)and \(\cos {\rm{(\pi - }}\phi ){\rm{ = - cos}}\phi \)

\(\iint_{S}{F}\cdot dS=\int_{0}^{2\pi }{\left( \int_{0}^{\pi }{1}6{{\sin }^{3}}\phi (-\cos \phi ){{\cos }^{2}}\theta -16{{\sin }^{3}}\pi {{\sin }^{2}}\theta +24{{\sin }^{2}}\phi (-\cos \phi )\cos \theta d\phi \right)}d\theta {{e}^{i\theta }}\)

\(\iint_{S}{F}\cdot dS=\int_{0}^{2\pi }{\left( \int_{0}^{\pi }{-}16{{\sin }^{3}}\phi \cos \phi {{\cos }^{2}}\theta -16{{\sin }^{3}}\pi {{\sin }^{2}}\theta -24{{\sin }^{2}}\phi \cos \phi \cos \theta d\phi \right)}d\theta \)

Add the red and blue equations

\(\begin{aligned}\iint_{S}{F}\cdot dS&=\int_{0}^{2\pi }{\left( \int_{0}^{\pi }{-}32{{\sin }^{3}}\pi {{\sin }^{2}}\theta d\phi \right)}d\theta \\ \iint_{S}{F}\cdot dS&=-16\left( \int_{0}^{2\pi }{{{\sin }^{2}}}\theta d\theta \right)\left( \int_{0}^{\pi }{{{\sin }^{3}}}\pi d\phi \right) \\ & =-16(A)(B).\end{aligned}\)

\({\rm{A = }}\int_{\rm{0}}^{{\rm{2\pi }}} {{\rm{si}}{{\rm{n}}^{\rm{2}}}} {\rm{\theta d\theta }}\)

Substitute \({\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta = }}\frac{{{\rm{1 - cos2\theta }}}}{{\rm{2}}}\)

\(\begin{array}{l}{\rm{A = }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{0}}^{{\rm{2\pi }}} {\rm{1}} {\rm{ - cos2\theta d\theta }}\\{\rm{A = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{\theta - }}\frac{{{\rm{sin2\theta }}}}{{\rm{2}}}} \right)_{\rm{0}}^{{\rm{2\pi }}}\end{array}\)

\({\rm{A = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ * 2\pi = \pi }}{\rm{.}}\)

\(\begin{array}{l}{\rm{B = }}\int_{\rm{0}}^{\rm{\pi }} {{\rm{si}}{{\rm{n}}^{\rm{3}}}} \phi d\phi \\{\rm{B = }}\int_0^{\rm{\pi }} {\left( {{\rm{si}}{{\rm{n}}^{\rm{2}}}\phi } \right)} {\rm{sin}}\phi d\phi \\{\rm{B = }}\int_0^{\rm{\pi }} {\left( {{\rm{1 - co}}{{\rm{s}}^{\rm{2}}}\phi } \right)} \sin \phi d\phi \end{array}\)

Substitute \({\rm{cos}}\phi {\rm{ = u}}\) and \({\rm{ - sin}}\phi d\phi {\rm{ = du}}\)

Step 4:Integration's limits will change.

Integration's limits will change from\(\int_{\rm{ - }} {{{\rm{0}}^{\rm{\pi }}}} {\rm{ to }}\mathop \smallint \nolimits_1^{ - 1} \)

\(\begin{array}{l}{\rm{B = }}\int_{\rm{1}}^{{\rm{ - 1}}} {{{\rm{u}}^{\rm{2}}}} {\rm{ - 1du}}\\{\rm{B = }}\left( {\frac{{{{\rm{u}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ - u}}} \right)_{\rm{1}}^{{\rm{ - 1}}}{\rm{ = }}\frac{{\rm{4}}}{{\rm{3}}}\end{array}\)

\(\iint_{S}{F}\cdot dS=-16(\pi )\left( \frac{4}{3} \right)=-\frac{64\pi }{3}.\)

The surface integral\(\iint_{S}{F}\cdot dS=-\frac{64\pi }{3}\).