Question

Use Stokes’ Theorem to evaluate \(\iint_{S}{curl}F\cdot dS,\)where \({\rm{F(x,y,z) = }}{{\rm{x}}^{\rm{2}}}{\rm{yzi + y}}{{\rm{z}}^{\rm{2}}}{\rm{j + }}{{\rm{z}}^{\rm{3}}}{{\rm{e}}^{{\rm{xy}}}}{\rm{k,S}}\) is the part of the sphere \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 5}}\) that lies above the plan \({\rm{Z = 1}}\), and \({\rm{S}}\) is oriented upward.

Short answer

Stokes’ Theoremevaluation \(\iint_{S}{curl}F\cdot dS=-4\pi \).

Step-by-step solution

Explanation of solution.

Stokes’ Theorem

\(\iint_{S}{curl}F\cdot dS=\int_{C}{F}\cdot dr\).

The boundary of the surface S is indicated by C.

Compute \(\int_{\rm{C}} {\rm{F}} \cdot d{\bf{r}}\)

The sphere and the plan \({\rm{z = 1}}\) intersect on a circle,

\({\rm{z = 1,}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 4}}\)

This can be written in parametric form as \({\rm{x = 2cost,y = 2sint,z = 1}}\)

Where \(0 < t \le {\rm{2\pi }}\)

\(\begin{aligned}\rm r &= \langle {\rm{2cost,2sint,1}}\rangle \\\rm dr &= \langle - {\rm{2sint,2cost,0}}\rangle dt\end{aligned}\)

Given

\({\rm{F = }}{{\rm{x}}^{\rm{2}}}{\rm{yzi + y}}{{\rm{z}}^{\rm{2}}}{\rm{j + }}{{\rm{z}}^{\rm{3}}}{{\rm{e}}^{{\rm{xy}}}}{\rm{k}}\).

Calculation.

Substitute the values of \({\rm{x,y}}\)and \({\rm{z}}\),

\({\rm{F = }}\left\langle {{\rm{8co}}{{\rm{s}}^{\rm{2}}}{\rm{t sin t,2sin t,}}{{\rm{e}}^{{\rm{4cost sint}}}}} \right\rangle \)

\(\begin{aligned}\int_{\rm{C}} {\rm{F}} \cdot \rm dr &= \int_0^{{\rm{2\pi }}} {\left\langle {{\rm{8co}}{{\rm{s}}^{\rm{2}}}{\rm{t sint,2sint,}}{{\rm{e}}^{{\rm{4costsint}}}}} \right\rangle } \cdot \langle {\rm{ - 2sint,2cost,0}}\rangle {\rm{dt}}\\\rm &= \int_{\rm{0}}^{{\rm{2\pi }}} {\rm{ - }} {\rm{16co}}{{\rm{s}}^{\rm{2}}}{\rm{t si}}{{\rm{n}}^{\rm{2}}}{\rm{t + 4cost sint dt}}\\\rm &= \int_{\rm{0}}^{{\rm{2\pi }}} {\rm{ - }} {\rm{4si}}{{\rm{n}}^{\rm{2}}}{\rm{2t + 2sin 2t dt}}\end{aligned}\)

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

\(\begin{array}{l}{\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\rm{2}} {\rm{(cos 4t - 1) + 2sin 2t dt}}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\rm{2}} {\rm{cos 4t + 2sin 2t - 2 dt}}\\{\rm{ = }}\left( {\frac{{{\rm{sin4t}}}}{{\rm{2}}}{\rm{ - cos2t - 2t}}} \right)_{\rm{0}}^{{\rm{2\pi }}}\\{\rm{ = - 4\pi }}{\rm{.}}\end{array}\)

Stokes’ Theorem evaluation \(\iint_{S}{curl}F\cdot dS=-4\pi \).