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. Compute the scalar suface integral Z Z K z dS where K is the part of the sphere of radius 3 centered at the origin, that lies in the first octant. (Hint: Use the parametrization g(θ, φ) = h3 sin φ cos θ, 3 sin φ sin θ, 3 cos φi with domain of definition [0, π/2] × [0, π/2])

Respuesta :

LammettHash

With the parameterization

[tex]\vec g(\theta,\varphi)=\langle3\sin\varphi,\cos\theta,3\sin\varphi,\sin\theta,3\cos\varphi\rangle[/tex]

take the normal vector to [tex]K[/tex] to be

[tex]\dfrac{\partial\vec g}{\partial\varphi}\times\dfrac{\partial\vec g}{\partial\theta}=\langle9\cos\theta\sin^2\varphi,9\sin\theta\sin^2\varphi,9\cos\varphi\sin\varphi\rangle[/tex]

which has magnitude

[tex]\left\|\dfrac{\partial\vec g}{\partial\varphi}\times\dfrac{\partial\vec g}{\partial\theta}\right\|=\sqrt{(9\cos\theta\sin^2\varphi)^2+(9\sin\theta\sin^2\varphi)^2+(9\cos\varphi\sin\varphi)^2}=9\sin\varphi[/tex]

Then the integral is

[tex]\displaystyle\iint_Kz\,\mathrm dS=\int_0^{\pi/2}\int_0^{\pi/2}3\cos\varphi(9\sin\varphi)\,\mathrm d\theta\,\mathrm d\varphi[/tex]

[tex]=\displaystyle\frac{27\pi}4\int_0^{\pi/2}\sin(2\varphi)\,\mathrm d\varphi=\boxed{\frac{27\pi}4}[/tex]

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