Stokes' theorem equates the surface integral of the curl of F to the line integral of F along the boundary of the hemisphere. The boundary itself is a circle C (the intersection of the hemisphere with the plane y = 0) with equation
[tex]x^2+z^2=16[/tex]
Parameterize this circle by
[tex]\mathbf r(t)=4\cos t\,\mathbf i+4\sin t\,\mathbf k[/tex]
with [tex]0\le t\le2\pi[/tex].
The surface is oriented such that its normal vector points in the positive y direction, which corresponds to the curve having counterclockwise orientation. The parameterization we're using here already takes this into account.
Now compute the line integral of F along C :
[tex]\displaystyle\iint_S\mathrm{curl}\mathbf F(x,y,z)\cdot\mathrm d\mathbf S=\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r[/tex]
[tex]=\displaystyle\int_0^{2\pi}\mathbf F(4\cos t,0,4\sin t)\cdot\frac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{2\pi}(4\sin t\,\mathbf i+4\cos t\,\mathbf j)\cdot(-4\sin t\,\mathbf i+4\cos t\,\mathbf k)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{2\pi}-16\sin^2t\,\mathrm dt[/tex]
[tex]=-8\displaystyle\int_0^{2\pi}(1-\cos(2t))\,\mathrm dt=\boxed{-16\pi}[/tex]
Line integral of F along C is,
[tex]\rm \int \int_S curl F(x,y,z) dS = -16\pi[/tex]
Step-by-step explanation:
Given :
Hemisphere - [tex]x^2 +y^2+z^2=16[/tex]
Calculation :
Accordind to Stoke's theorem the surface integral of the curl of F to the line integral of F along the boundary of the hemisphere. The boundary itself is a circle C (the intersection of the hemisphere with the plane y = 0) with equation
[tex]x^2+z^2=16[/tex]
then parameterize the circle,
[tex]\rm r(t) = 4 cos(t) \;\hat{i} + 4 sin(t)\;(\hat{k})[/tex]
with , [tex]0\leq t\leq 2\pi[/tex]
Line integral of F along C is,
[tex]\rm \int \int_S curl F(x,y,z) dS = \int_{C}^{} F(x,y,z) \;dr[/tex]
[tex]= \int_{0}^{2\pi} F(4cos(t),0,4sin(t)) \;\dfrac{dr}{dt}.dt[/tex]
[tex]= \int_{0}^{2\pi}(4sin(t)i+4cos(t) j).(-4sin(t)i+4cos(t)k) \;dt[/tex]
[tex]= \int_{0}^{2\pi} -16sin^2tdt[/tex]
[tex]=-8 \int_{0}^{2\pi} (1-cos(2t))dt[/tex]
[tex]= -16\pi[/tex]
For more information, refer the link given below
https://brainly.com/question/8130922?referrer=searchResults
