Question

Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use. F = <<x sin y, -cos y, z sin y>> , S is the boundary of the region bounded by the planes x = 1, y = 0, y = pi/2, z = 0, and z = x.

Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use.
\(\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{\sin{{y}}}\right\rangle}\) , S is the boundary of the region bounded by the planes x = 1, y = 0, \(\displaystyle{y}=\frac{\pi}{{2}},{z}={0}\), and z = x.

Answers (1)

2021-02-01
Step 1
Given:
\(\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{z}{\sin{{y}}}\right\rangle}\), the surface S is the boundary of the region bounded by the planes x = 1, \(\displaystyle{y}=\frac{\pi}{{2}},{z}={0}\), and z = x.
Step 2
Using Divergence theorem
\(\displaystyle\int\int_{{D}}\int\nabla\cdot{F}{d}{v}=\int\int_{{S}}{F}\cdot{n}{d}{s}\)
Compute The divergence of the field
\(\displaystyle\nabla\cdot{F}=\frac{\partial}{{\partial{x}}}{\left({x}{\sin{{y}}}\right)}+\frac{\partial}{{\partial{y}}}{\left(-{\cos{{y}}}\right)}+\frac{\partial}{{\partial{z}}}{\left({z}{\sin{{y}}}\right)}\)
\(\displaystyle={\sin{{y}}}\frac{\partial}{{\partial{x}}}{\left({x}\right)}+\frac{\partial}{{\partial{y}}}{\left(-{\cos{{y}}}\right)}+{\sin{{y}}}\frac{\partial}{{\partial{z}}}{\left({z}\right)}\)
\(\displaystyle={\sin{{y}}}+{\sin{{y}}}+{\sin{{y}}}\)
\(\displaystyle={3}{\sin{{y}}}\)
Step 3
Therefore, the outward flux is
\(\displaystyle\int\int_{{D}}\int\nabla\cdot{F}{d}{v}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{1}}}{\int_{{0}}^{{x}}}{3}{\sin{{y}}}{\left.{d}{z}\right.}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={3}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{1}}}{\sin{{y}}}{{\left[{z}\right]}_{{0}}^{{x}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={3}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{1}}}{x}{\sin{{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={3}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\left[\frac{{x}^{{2}}}{{2}}\right]}_{{0}}^{{1}}}{\sin{{y}}}{\left.{d}{y}\right.}\)
\(\displaystyle=\frac{{3}}{{2}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left[{\sin{{y}}}{\left.{d}{y}\right.}\right.}\)
\(\displaystyle=\frac{{3}}{{2}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left[{\sin{{y}}}{\left.{d}{y}\right.}\right.}\)
\(\displaystyle=\frac{{3}}{{2}}{{\left[-{\cos{{y}}}\right]}_{{0}}^{{\frac{\pi}{{2}}}}}\)
\(\displaystyle=\frac{{3}}{{2}}{\left(-{\cos{{\left(\frac{\pi}{{2}}\right)}}}-{\left(-{\cos{{\left({0}\right)}}}\right)}\right)}\)
\(\displaystyle=\frac{{3}}{{2}}{\left({0}-{\left(-{1}\right)}\right)}\)
\(\displaystyle=\frac{{3}}{{2}}\)
Step 4
Thus, the outward flux of the given field is \(\displaystyle\frac{{3}}{{2}}\).
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