Respuesta :
Parameterize the part of the sphere you're interested in, [tex]\mathcal S[/tex], by
[tex]\vec s(u,v)=\underbrace{a\cos u\sin v}_{x(u,v)}\,\vec\imath+\underbrace{a\sin u\sin v}_{y(u,v)}\,\vec\jmath+\underbrace{a\cos v}_{z(u,v)}\,\vec k[/tex]
with [tex]0\le u\le\pi[/tex] and [tex]0\le v\le\pi[/tex]. The domain of [tex]u[/tex] guarantees that we're dealing with only positive values of [tex]y(u,v)[/tex].
Then the flux of [tex]\vec F=3z\,\vec k[/tex] over [tex]\mathcal S[/tex] is given by the surface integral
[tex]\displaystyle\iint_{\mathcal S}(\vec F\cdot\vec n)\,\mathrm d\sigma=\int_{v=0}^{v=\pi}\int_{u=0}^{u=\pi}\vec F(\vec s(u,v))\cdot\left(\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right)\,\mathrm du\,\mathrm dv[/tex]
where the order of the partial derivatives is chosen to make sure we pick the normal vector to [tex]x^2+y^2+z^2=a^2[/tex] pointing in the positive [tex]y[/tex] direction.
We have
[tex]\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=a^2\left(\cos u\sin^2v\,\vec\imath+\sin u\sin^2v\,\vec\jmath+\cos v\sin v\,\vec k\right)[/tex]
[tex]\vec F(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right)=a^3\cos^2v\sin v[/tex]
Then the integral reduces to
[tex]\displaystyle a^3\int_{v=0}^{v=\pi}\int_{u=0}^{u=\pi}\cos^2v\sin v\,\mathrm du\,\mathrm dv[/tex]
[tex]\displaystyle a^3\pi\int_{v=0}^{v=\pi}\cos^2v\sin v\,\mathrm dv[/tex]
Substitute [tex]w=\cos v[/tex], [tex]\mathrm dw=-\sin v\,\mathrm dv[/tex], then
[tex]\displaystyle-a^3\pi\int_{w=1}^{w=-1}w^2\,\mathrm dw[/tex]
[tex]\displaystyle a^3\pi\int_{w=-1}^{w=1}w^2\,\mathrm dw[/tex]
[tex]\displaystyle2a^3\pi\int_{w=0}^{w=1}w^2\,\mathrm dw[/tex]
[tex]2a^3\pi\dfrac{w^3}3\bigg|_{w=0}^{w=1}[/tex]
which leaves us with a flux of [tex]\dfrac{2a^3\pi}3[/tex].
Following are the calculation to the given qestion:
Consider the vector function:
[tex]\to F(x,y,z)=3zk = (0,0,3z)[/tex]
Normal vector:
[tex]\to \overrightarrow{n} = \frac{(xi, yj+zk)}{a} =\frac{(x,y,z)}{a}[/tex]
When the dot product of vectors= [tex]F,\overrightarrow{n}[/tex] :
[tex]\to F \cdot \overrightarrow{n} =(0,0,3z) \cdot \frac{(x,y,z)}{a}=\frac{3z^2}{a}[/tex]
Using sphereical coordinates,
[tex]\to x=a \cos \theta \sin \phi\\\\ \to y= a \sin \theta \sin \phi \\\\ \to z = a \cos \phi[/tex]
So, the dot product is
[tex]\to F \cdot \overrightarrow{n}=\frac{ 3z^2}{a} =\frac{3(a \cos \phi)^2}{a}[/tex]
[tex]=\frac{3a^2 \cos^2 \phi}{a} \\\\ = 3a \cos^2 \phi[/tex]
The area element on the surface of the sphere of radius:
[tex]\to r = a \ is\ (ad \theta) (a \sin \phi d \phi) = a^2 sin \phi d \theta d \phi\\\\[/tex]
Now, the surface integral:
[tex]\to \int \int_{S} F \cdot \ dS = \int^{\frac{\pi}{2}}_{\phi \to 0} \int^{2\pi}_{d \to 0} (3a cos^2 \phi)a^2 \sin \phi d \theta d \phi \\\\[/tex]
[tex]= 3a^3 \int^{\frac{\pi}{2}}_{\phi \to 0} \int^{2\pi}_{d \to 0} ( cos^2 \phi) \sin \phi d \theta d \phi \\\\= 3a^3 \int^{\frac{\pi}{2}}_{\phi \to 0} ([( - \frac{cos^3 \phi}{3})])^{2\pi}_{d \to 0} \ d \theta \\\\= 3a^3 \int^{\frac{\pi}{2}}_{\phi \to 0} ([( - \frac{cos^3 ({\frac{\pi}{2}}))} {3} + \frac{\cos^3 \ (0)}{3})]) \ d \theta \\\\= 3a^3 \int^{\frac{\pi}{2}}_{\phi \to 0} ([( - \frac{0}{3} + \frac{1}{3})]) \ d \theta \\\\= 3a^3 \int^{\frac{\pi}{2}}_{\phi \to 0} ([ \frac{1}{3}]) \ d \theta \\\\[/tex]
[tex]= a^3 \int^{\frac{\pi}{2}}_{\phi \to 0} (1) \ d \theta \\\\=a^3 (2\pi)\\\\=2\pi a^3\\\\[/tex]
Therefore, the required flux is [tex]\int \int_{S}\ F.dS = 2\pi a^3[/tex]
Learn more:
brainly.com/question/6929972
