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Use Stoke's Theorem to evaluate∬M(∇×F)⋅dS where M is the hemisphere x2+y2+z2=16,x≥0, with the normal in the direction of the positive x direction, and F=⟨x2,0,y2⟩. Begin by writing down the "standard" parametrization of ∂M as a function of the angle θ (denoted by "t" in your answer) x= , y=

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LammettHash

[tex]\partial M[/tex] is the circle with radius 4 centered at (0, 0, 0) lying in the plane [tex]x=0[/tex]. We can parameterize it by

[tex]\vec r(t)=\langle0,4\cos t,4\sin t\rangle[/tex]

with [tex]0\le t\le2\pi[/tex]. By Stokes' theorem,

[tex]\displaystyle\iint_M(\nabla\times\vec F)\cdot\mathrm d\vec S=\int_{\partial M}\vec F\cdot\mathrm d\vec r[/tex]

[tex]=\displaystyle\int_0^{2\pi}\langle0^2,0,16\cos^2t\rangle\cdot\langle0,-4\sin t,4\cos t\rangle\,\mathrm dt[/tex]

[tex]=\displaystyle\int_0^{2\pi}64\cos^3t\,\mathrm dt=\boxed0[/tex]

ajeigbeibraheem

Using Stoke's Theorem, the value of the integral is zero = 0

According to Stoke's theorem, the surface integral of the curl of a function above a surface bounded by a closed circular surface is equivalent to the line integral of the specific vector function about that surface.

Considering the following vector field;

  • [tex]\mathbf{F = \langle x^2, 0, y^2 \rangle}[/tex] and using,
  • Stoke's Theorem [tex]\mathbf{\iint_S curl F^\to dS^\to = \oint _CF.dr}[/tex]

Here;

  • M is the hemisphere x²+y²+z² = 16, x ≥ 0 oriented upward with the normal in the direction of the positive x-axis.

Thus, the projection on yz-plane is circle and x = 0

∴

y²+z² = 16

The standard parametrization of the circle can be expressed as:

x = 0, y = 4 cos θ, z = 4 sin θ

[tex]\mathbf{M = \langle 0, 4cos \theta , 4 sin \theta\rangle}[/tex]

[tex]\mathbf{\partial M = \langle 0, -4sin \theta , 4cos \theta\rangle}[/tex]

∴

[tex]\mathbf{F(M) = F \langle 0, 4cos \theta , 4sin \theta\rangle}[/tex]

[tex]\mathbf{F(M) = \langle 0^2, 0 ,(4cos \theta)^2 \rangle}[/tex]

[tex]\mathbf{F(M) = \langle 0, 0 ,16cos^2 \theta \rangle}[/tex]

  • Since θ varies from 0 to 2π

By Stoke's Theorem, the integral can be computed as:

[tex]\mathbf{\int \partial MF.dS = \int ^{2\pi}_{0} \langle 0,0,16 cos^2 \theta \rangle \langle 0, -4sin \theta, 4 cos \theta\rangle d \theta}[/tex]

[tex]\mathbf{\implies \int^{2 \pi}_{0} ( 0+0+64 cos ^3 \theta) d \theta}[/tex]

[tex]\mathbf{\implies \int^{2 \pi}_{0} (64 cos ^3 \theta) d \theta}[/tex]

[tex]\mathbf{\implies 64 \int^{2 \pi}_{0} ( cos ^2 \theta)cos \theta d \theta}[/tex]

[tex]\mathbf{\implies 64 \int^{2 \pi}_{0} (1 - sin^2 \theta) cos \theta d \theta}[/tex]

[tex]\mathbf{\implies 64 \int^{2 \pi}_{0} (cos \theta - cos \theta sin^2 \theta) d \theta}[/tex]

Suppose:

  • u = sin x
  • du = cos x dx
  • x = 0, u = 0
  • x = 2π, u = 0

∴

[tex]\mathbf{\implies 64 \Bigg [[sin \theta]^{2 \pi}_{0}- \int ^0_0 u^2 du \Bigg ] }[/tex]

⇒ 64 [0 - 0]

⇒ 0

Therefore, we can conclude that the value of the integral is 0

Learn more about Stoke's Theorem here:

https://brainly.com/question/13051503?referrer=searchResults

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