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Sagot :
The flux of out of S is [tex]14 \pi\left(-\frac{\sqrt{2}}{2}+1\right)[/tex].
What is Flux of the Vector Field?
Solving the flux of the vector field using the divergence theorem which has the following formula [tex]$\iint_S F \cdot d S=\iiint_E div} F(x, y, z) d V$[/tex] where F is the vector field [tex]$F=P i+Q j+R k$[/tex] and div F means the divergence of F which has the formula [tex]$F=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$[/tex]
Here, we let the vector field to be [tex]$\vec{F}=x i+y j+z k$[/tex]
[tex]$\vec{F}=x i+y j+z k$[/tex]
Now solving for [tex]${div} F$[/tex]
[tex]$\begin{aligned}& {div} F=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z} \\& {div} F=1+1+1=3\end{aligned}$[/tex]
Thus,
[tex]\iint_S F \cdot d S=\iiint 3 d V[/tex]
From the given region the limits of integrations in spherical coordinates are,
[tex]0 \leq \theta \leq 2 \pi, 0 \leq \phi \leq \frac{\pi}{4}, 1 \leq \rho \leq 2[/tex]
Thus,
[tex]\iint_S F \cdot d S=\int_0^{2 \pi} \int_0^{\frac{\pi}{4}} \int_1^2 3 \rho^2 \sin \phi d \rho d \phi d \theta[/tex]
Integrate with respect to [tex]$\rho$[/tex]
[tex]\begin{aligned}& =\int_0^{2 \pi} \int_0^{\frac{\pi}{4}} 3\left[\frac{1}{3} \rho^3\right]_1^2 \sin \phi d \phi d \theta \\& =\int_0^{2 \pi} \int_0^{\frac{\pi}{4}} 7 \sin \phi d \phi d \theta\end{aligned}[/tex]
Integrate with respect to [tex]$\phi$[/tex]
[tex]\begin{aligned}& =\int_0^{2 \pi} 7[-\cos \phi]_0^{\frac{\pi}{4}} d \theta \\& =\int_0^{2 \pi} 7\left(-\frac{\sqrt{2}}{2}+1\right) d \theta\end{aligned}[/tex]
Integrate with respect to [tex]$\theta$[/tex]
[tex]\begin{aligned}& =7\left(-\frac{\sqrt{2}}{2}+1\right)[\theta]_0^{2 \pi} \\& =14 \pi\left(-\frac{\sqrt{2}}{2}+1\right)\end{aligned}[/tex]
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