Answer to Question #110371 in Calculus for kian

Question

Evaluate the ∫∫ z dS where S is the upper half of a sphere of radius 4.

Answer 1

We use the spherical coordinates

$x=r \sin \varphi \cos \theta,\ \ \ y=r \sin \varphi \sin \theta ,\ \ \ z= \cos \varphi$

Since $r=4$ , then the parameterization for this sphere

$\vec{r}(\theta, \varphi)=4 \sin \varphi \cos \theta \vec{\imath}+4 \sin \varphi \sin \theta \vec{\jmath}+4 \cos \varphi \vec{k}$

Since we are working on the upper half of the sphere here are the limits on the parameters.

$0 \leq \theta \leq 2 \pi \quad 0 \leq \varphi \leq \frac{\pi}{2}$

Since

$\begin{array}{l} \vec{r}_{\theta}(\theta, \varphi)=-2 \sin \varphi \sin \theta \vec{i}+2 \sin \varphi \cos \theta \vec{j} \\ \vec{r}_{\varphi}(\theta, \varphi)=2 \cos \varphi \cos \theta \vec{i}+2 \cos \varphi \sin \theta \vec{j}-2 \sin \varphi \vec{k} \end{array}$

Then

$\begin{aligned} \vec{r}_{\theta} \times \vec{r}_{\varphi} &=\left|\begin{array}{cc} \vec{i} & \vec{j} & \vec{k} \\ -4\sin \varphi \sin \theta & 4 \sin \varphi \cos \theta & 0 \\ 4 \cos \varphi \cos \theta & 4 \cos \varphi \sin \theta & -4\sin \varphi \end{array}\right| \\ &=-16 \sin ^{2} \varphi \cos \theta \vec{i}-16 \sin \varphi \cos \varphi \sin ^{2} \theta \vec{k}-16 \sin ^{2} \varphi \sin \theta \vec{j}\\ &\ \ \ \ \ \ -16 \sin \varphi \cos \varphi \cos ^{2} \theta \vec{k} \\ &=-16 \sin ^{2} \varphi \cos \theta \vec{i}-16 \sin ^{2} \varphi \sin \theta \vec{j}\\ &\ \ \ \ \ -16 \sin \varphi \cos \varphi\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \vec{k} \\ &=-16 \sin ^{2} \varphi \cos \theta \vec{i}-16 \sin ^{2} \varphi \sin \theta \vec{j} -16 \sin \varphi \cos \varphi \vec{k} \end{aligned}$

Hence the magnitude

$\begin{aligned} \left\|\vec{r}_{\theta} \times \vec{r}_{\varphi}\right\| &=\sqrt{16^2 \sin ^{4} \varphi \cos ^{2} \theta+16^2 \sin ^{4} \varphi \sin ^{2} \theta+16 ^2\sin ^{2} \varphi \cos ^{2} \varphi} \\ &=\sqrt{16^2 \sin ^{4} \varphi\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+16^2 \sin ^{2} \varphi \cos ^{2} \varphi} \\ &=\sqrt{16^2 \sin ^{2} \varphi\left(\sin ^{2} \varphi+\cos ^{2} \varphi\right)}\\ &=16\sin \varphi \end{aligned}$

The surface integral is 

$\begin{aligned} \iint\limits_{S}{{z\,dS}} &= \iint\limits_{D}{{4\cos \varphi \left( {16\sin \varphi } \right)\,dA}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,\frac{\pi }{2}}}{{32\sin \left( {2\varphi } \right)\,d\varphi }}\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{ {\left( { - 16\cos \left( {2\varphi } \right)} \right)} \bigg|_0^{\frac{\pi }{2}}\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{16\,d\theta }}\\ & = 32\pi \end{aligned}$