Parameterize [tex]S[/tex] by
[tex]\vec s(u,v)=9\cos u\sin v\,\vec\imath+9\sin u\sin v\,\vec\jmath+9\cos v\,\vec k[/tex]
with [tex]0\le u\le\dfrac\pi2[/tex] and [tex]0\le v\le\dfrac\pi2[/tex]. Take the normal vector to [tex]S[/tex] to be
[tex]\vec s_u\times\vec s_v=-81\cos u\sin^2v\,\vec\imath-81\sin u\sin^2v\,\vec\jmath-81\cos v\sin v\,\vec k[/tex]
Then the flux of [tex]\vec F[/tex] across [tex]S[/tex] with the given orientation is
[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S[/tex]
[tex]\displaystyle=\int_0^{\pi/2}\int_0^{\pi/2}\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv[/tex]
where
[tex]\vec F(x(u,v),y(u,v),z(u,v))=9\cos u\sin v\,\vec\imath-9\cos v\,\vec\jmath+9\sin u\sin v\,\vec k[/tex]
The integral reduces to
[tex]\displaystyle-729\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv=\boxed{-\frac{243\pi}2}[/tex]
Surface integral involves combining numerous integrals to integrate over different surfaces. The evaluation of the surface integral is: [tex]\frac{243}{2} \pi[/tex]
Given that:
[tex]F(x, y, z) = x_i - z_j + y_k[/tex] and [tex]x^2 + y^2 + z^2 = 81[/tex]
Differentiate each term of [tex]F(x, y, z) = x_i - z_j + y_k[/tex]
[tex]dF = \frac{df_1}{dx} + \frac{df_2}{dz} + \frac{df_3}{dy}[/tex]
[tex]dF = \frac{d}{dx}(x) + \frac{d}{dz}(-z) + \frac{d}{dy}(y)[/tex]
The differentiation for [tex]x_i[/tex] will be 1, while others will be 0.
So, the equation becomes
[tex]dF = 1 + 0 + 0[/tex]
[tex]dF = 1[/tex]
Next calculate the radius of the sphere:
We have:
[tex]x^2 + y^2 + z^2 = 81[/tex]
Express 81 as [tex]9^2[/tex]
[tex]x^2 + y^2 + z^2 = 9^2[/tex]
A sphere is represented as:
[tex]x^2 + y^2 + z^2 = r^2[/tex]
Where:
[tex]r \to[/tex] radius
By comparison
[tex]r^2 = 9^2[/tex]
Take positive square roots
[tex]r = 9[/tex]
The volume (V) is then calculated as:
[tex]V = \int\limits^{}_{v}dF \times dV[/tex]
Substitute [tex]dF = 1[/tex]
[tex]V = \int\limits^{}_{v}1 \times dV[/tex]
[tex]V = \int\limits^{}_{v} dV[/tex]
From the question, we understand that the sphere is in the first octant.
The first octant is represented as [tex]\frac{1}{8}[/tex]
The integral of [tex]V = \int\limits^{}_{v} dV[/tex] becomes
[tex]V = \frac{1}{8} \times v[/tex]
Where:
[tex]v = \frac{4}{3}\pi r^3[/tex] --- volume of a sphere
So, the equation becomes:
[tex]V = \frac 18 \times \frac{4}{3}\pi r^3[/tex]
Substitute [tex]r = 9[/tex]
[tex]V = \frac 18 \times \frac{4}{3}\pi \times 9^3[/tex]
[tex]V = \frac 18 \times \frac{4}{3}\pi \times 729[/tex]
[tex]V = \frac 18 \times 4\pi \times 243[/tex]
[tex]V = \frac 12 \times \pi \times 243[/tex]
[tex]V = \frac{243}{2} \pi[/tex]
Hence, the evaluation of the surface integral is: [tex]\frac{243}{2} \pi[/tex]
Read more about surface integral at:
https://brainly.com/question/11722148