Question

To determine the value of for \({\rm{F}}(x,y,z) = xzi + xj + y{\rm{k}}\) is \(0\).

Short answer

The value of for \({\rm{F}}(x,y,z) = xz{\rm{i}} + x{\rm{j}} + y{\rm{k}}\) is \(0\).

Step-by-step solution

Concept of Integrals

An integral assigns numbers to functions in a way that describes displacement, area, volume. The definite integral (also called Riemann integral) of a function\(f(x)\)is denoted as

\(\int_a^b f (x)dx\)

Calculation of the cross product\({r_\theta } \times {r_\phi }\)

The given function is \({\rm{F}}(x,y,z) = xz{\rm{i}} + x{\rm{j}} + y{\rm{k}}\) and the given curve is \({x^2} + {y^2} + {z^2} = 25\).

The parameterization of the given hemisphere

\(r(\theta ,\phi ) = 5\sin \phi \cos \theta {\bf{i}} + 5\sin \phi \sin \theta {\bf{j}} + 5\cos \phi {\bf{k}}\)

Where, \(\theta \in (0,\pi )\) and \(\phi \in (0,\pi )\).

\({r_\theta } = - 5\sin \phi \sin \theta {\bf{i}} + 5\sin \phi \cos \theta {\bf{j}}\)

\({r_\phi } = 5\cos \phi \cos \theta {\bf{i}} + 5\cos \phi \sin \theta {\bf{j}} - 5\sin \phi {\bf{k}}\)

The cross product \({r_\theta } \times {r_\phi }\) is obtained as follows.

\(\begin{array}{l}{r_\theta } \times {r_\phi } = \left| {\begin{array}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{ - 5\sin \phi \sin \theta }&{5\sin \phi \cos \theta }&0\\{5\sin \phi \cos \theta }&{5\cos \phi \sin \theta }&{ - 5\sin \phi }\end{array}} \right|\\{r_\theta } \times {r_\phi } = {\bf{i}}\left| {\begin{array}{*{20}{c}}{5\sin \phi \cos \theta }&0\\{5\cos \phi \sin \theta }&{ - 5\sin \phi }\end{array}} \right| - {\bf{j}}\left| {\begin{array}{*{20}{c}}{5\sin \phi \sin \theta }&0\\{5\cos \phi \cos \theta }&{ - 5\sin \phi }\end{array}} \right| + {\bf{k}}\left| {\begin{array}{*{20}{c}}{ - 5\sin \phi \sin \theta }&{5\sin \phi \cos \theta }\\{5\cos \phi \cos \theta }&{5\cos \phi \sin \theta }\end{array}} \right|\\{r_\theta } \times {r_\phi } = - 25{\sin ^2}\phi \cos \theta {\bf{i}} - 25{\sin ^2}\phi \sin \theta {\bf{j}} - 25\sin \phi \cos \phi {\bf{k}}\end{array}\)

Hence\(dS\) is given as follows.

\(\begin{array}{l}dS = - {\left( {{r_\theta } \times r} \right)_\phi }dA\\dS = \left( {25{{\sin }^2}\phi \cos \theta {\bf{i}} + 25{{\sin }^2}\phi \sin \theta {\bf{j}} + 25\sin \phi \cos \phi {\bf{k}}} \right)dA\end{array}\)

Calculation of the integral

It is given that \({\rm{F}}(x,y,z) = xz{\rm{i}} - z{\rm{j}} + y{\rm{k}}\). Therefore, \({\bf{F}}(r(\theta ,\phi ))\) is obtained as follows. \({\bf{F}}(r(\theta ,\phi )) = 25\cos \theta \sin \phi \cos \phi {\bf{i}} + 5\cos \theta \sin \phi {\bf{j}} + 5\sin \theta \sin \phi {\bf{k}}\)

Obtain the value of flux as follows.

Replace \(\sin (\pi - \theta ) = \sin \theta \) and \(\cos (\pi - \theta ) = - \cos \theta \) by substituting \(\pi - \phi \) in place of \((\pi - \theta )\) and \(\pi \) and obtain the flux as follows.

Add the flux in \(\left( 1 \right)\)and \(\left( 2 \right),\)and obtain the total flux as follows.

Therefore, the value of for \({\rm{F}}(x,y,z) = xz{\rm{i}} + x{\rm{j}} + y{\rm{k}}\) is \(0\).