Question

. 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])

Answer 1

With the parameterization

`\vec g(\theta,\varphi)=\langle3\sin\varphi,\cos\theta,3\sin\varphi,\sin\theta,3\cos\varphi\rangle`

take the normal vector to
`K` to be

`(\partial\vec g)/(\partial\varphi)*(\partial\vec g)/(\partial\theta)=\langle9\cos\theta\sin^2\varphi,9\sin\theta\sin^2\varphi,9\cos\varphi\sin\varphi\rangle`

which has magnitude

`\left\|(\partial\vec g)/(\partial\varphi)*(\partial\vec g)/(\partial\theta)\right\|=√((9\cos\theta\sin^2\varphi)^2+(9\sin\theta\sin^2\varphi)^2+(9\cos\varphi\sin\varphi)^2)=9\sin\varphi`

Then the integral is

`\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`

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