Respuesta :
Parameterize [tex]S[/tex] in cylindrical coordinates by
[tex]\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k[/tex]
where
[tex]\begin{cases}x(u,v)=u\cos v\\y(u,v)=x(u,v)^2+z(u,v)^2=u^2\\z(u,v)=u\sin v\end{cases}[/tex]
with [tex]0\le u\le2[/tex] and [tex]0\le v\le2\pi[/tex].
Take the normal vector to [tex]S[/tex] to be
[tex]\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath+u\,\vec\jmath-2u^2\sin v\,\vec k[/tex]
Then the flux of [tex]\vec F[/tex] across [tex]S[/tex] is
[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^2(3u^2\,\vec\imath+4\,\vec\jmath-3u^2\cos v\sin v\,\vec k)\cdot\left(\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_0^{2\pi}\int_0^2(4u-6u^4\cos^3v)\,\mathrm du\,\mathrm dv=\boxed{16\pi}[/tex]
The required flux is [tex]16\pi[/tex].
Given equation is,
[tex]\overrightarrow{F}=3y\hat{i} +4\hat{j}-3xz\hat{k}[/tex] through the surface [tex]S[/tex] bounded by [tex]y=x^{2} +z^{2}[/tex] with [tex]x^{2} +z^2\leq 4[/tex]
Computation:
Since, [tex]y=x^2+z^2[/tex] we have [tex]y_x=2z[/tex]
So,
[tex]d\overleftarrow{A}=(-y_x\hat{i}+\hat{j}-y_z\hat{kdxdz})\\=(-2x\hat{i}+\hat{j}-2z\hat{k})dxdz[/tex]
Then the flux will be,
[tex]\iint_{s}^{}F.ds= \iint_{S}^{} \left<3y,4,-3xz \right>\left<-2x,1,-2z \right>dA\\=\iint_{s}^{}(-6xy+4+6xz^2) dA\\=\iint_{x^2+z^2\leq 4}^{} (-6x(x^2+z^2)+4+6xz^2)dxdz[/tex]
The region bounded by [tex]x^{2}+y^{2} \leq 4[/tex].
So converting [tex]x[/tex],[tex]y[/tex] into polar co-ordinates we have [tex]x=rcos\theta , z=rsin\theta[/tex]
The region [tex]0\leq r\leq 2[/tex] and [tex]0\leq \theta \leq 2\pi[/tex]
Therefore,
[tex](-6x(x^2+z^2)+4+6xz^2)=-6rcos\theta (r^2cos^2+r^2sin^2\theta )+4+6rcos\theta r^2sin^2\\=-6r^3cos\theta +4+6r^3cos\theta sin^2\theta[/tex]
Then,
[tex]\iint_{s}^{}F.dS=\int_{0}^{2\pi }\int_{0}^{2\pi }(-6r^3cos\theta +4+6r^3cos\theta sin^2\theta )rdrd\theta \\=\int_{0}^{2\pi }[\frac{-6r^5cos\theta }{5}+\frac{4r^2}{2}+\frac{6r^5cos\theta sin^2\theta }{5}]^2_0d\theta \\=[\frac{-192sin\theta }{5}+\frac{16\theta }{2}+\frac{192sin^3\theta }{15}]^\pi_0 \\=16\pi[/tex]
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