Parameterize the surface (call it [tex]S[/tex]) that has [tex]C[/tex] as its boundary by
[tex]\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(81-u^2)\,\vec k[/tex]
with [tex]0\le u\le9[/tex] and [tex]0\le v\le2\pi[/tex].
Take the normal vector to [tex]S[/tex] to be
[tex]\vec n=\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k[/tex]
Compute the curl of [tex]\vec F(x(u,v),y(u,v),z(u,v))=\vec F(u,v)[/tex]. We have
[tex]\nabla\times\vec F(x,y,z)=4y\,\vec\imath-4x\,\vec\jmath[/tex]
[tex]\implies\nabla\times\vec F(u,v)=4u\sin v\,\vec\imath-4u\cos v\,\vec\jmath[/tex]
Then by Stoke's theorem, the line integral is
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F(x,y,z))\cdot\mathrm d\vec S[/tex]
[tex]=\displaystyle\iint_S(\nabla\times\vec F(u,v))\cdot\vec n\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_0^{2\pi}\int_0^90\,\mathrm du\,\mathrm dv=\boxed{0}[/tex]
We can verify this result by computing the line integral directly. Parameterize [tex]C[/tex] by
[tex]\vec r(t)=9\cos t\,\vec\imath+9\sin t\,\vec\jmath[/tex]
with [tex]0\le t\le2\pi[/tex]. Then
[tex]\displaystyle\int_C\vec F(x,y,z)\cdot\mathrm d\vec r=\int_0^{2\pi}\vec F(t)\cdot\frac{\mathrm d\vec r}{\mathrm dt}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{2\pi}(9\cos t\,\vec\imath+9\sin t\,\vec\jmath+162\,\vec k)\cdot(-9\sin t\,\vec\imath+9\cos t\,\vec\jmath)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{2\pi}0\,\mathrm dt=\boxed{0}[/tex]
By applying Stokes theorem, the value of the integral is equal to zero (0).
First of all, we would use Stoke's theorem to convert the integral as follows:
[tex]\int_s(\Delta \times F).mds = \int_c Fdr[/tex]
where:
c represents the boundary circle on the xy-plane.
We know that F(x, y, z) = xi + yj + 2(x² + y²)k. Thus, we would compute the curl of F by using a matrix:
[tex]\Delta \times F = \left[\begin{array}{ccc}i&j&k\\d/dx&d/dy&d/dz\\x&y&2(x^2 +y^2)\end{array}\right][/tex]
Δ × F = i(4y) - j(4x)
Δ × F = 4yî - 4j
By taking the normal vector of S, we have:
S; x² + y² + z² - 81 = 0
ds = 2xî + 2yj + 2zk
By applying Stoke's theorem, the line integral is given by:
(Δ × F) . ds = (4yî - 4j) . (2xî + 2yj + 2zk)
(Δ × F) . ds = 8xy - 8xy + 0
(Δ × F) . ds = 0.
∫cF⋅dr = 0.
Read more on Stoke's theorem here: https://brainly.com/question/13105453
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