We use the spherical coordinates
x = r sin φ cos θ , y = r sin φ sin θ , z = cos φ x=r \sin \varphi \cos \theta,\ \ \ y=r \sin \varphi \sin \theta ,\ \ \ z= \cos \varphi x = r sin φ cos θ , y = r sin φ sin θ , z = cos φ
Since r = 4 r=4 r = 4
r ⃗ ( θ , φ ) = 4 sin φ cos θ ı ⃗ + 4 sin φ sin θ ȷ ⃗ + 4 cos φ k ⃗ \vec{r}(\theta, \varphi)=4 \sin \varphi \cos \theta \vec{\imath}+4 \sin \varphi \sin \theta \vec{\jmath}+4 \cos \varphi \vec{k} r ( θ , φ ) = 4 sin φ cos θ + 4 sin φ sin θ + 4 cos φ k
Since we are working on the upper half of the sphere here are the limits on the parameters.
0 ≤ θ ≤ 2 π 0 ≤ φ ≤ π 2 0 \leq \theta \leq 2 \pi \quad 0 \leq \varphi \leq \frac{\pi}{2} 0 ≤ θ ≤ 2 π 0 ≤ φ ≤ 2 π
Since
r ⃗ θ ( θ , φ ) = − 2 sin φ sin θ i ⃗ + 2 sin φ cos θ j ⃗ r ⃗ φ ( θ , φ ) = 2 cos φ cos θ i ⃗ + 2 cos φ sin θ j ⃗ − 2 sin φ k ⃗ \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} r θ ( θ , φ ) = − 2 sin φ sin θ i + 2 sin φ cos θ j r φ ( θ , φ ) = 2 cos φ cos θ i + 2 cos φ sin θ j − 2 sin φ k
Then
r ⃗ θ × r ⃗ φ = ∣ i ⃗ j ⃗ k ⃗ − 4 sin φ sin θ 4 sin φ cos θ 0 4 cos φ cos θ 4 cos φ sin θ − 4 sin φ ∣ = − 16 sin 2 φ cos θ i ⃗ − 16 sin φ cos φ sin 2 θ k ⃗ − 16 sin 2 φ sin θ j ⃗ − 16 sin φ cos φ cos 2 θ k ⃗ = − 16 sin 2 φ cos θ i ⃗ − 16 sin 2 φ sin θ j ⃗ − 16 sin φ cos φ ( sin 2 θ + cos 2 θ ) k ⃗ = − 16 sin 2 φ cos θ i ⃗ − 16 sin 2 φ sin θ j ⃗ − 16 sin φ cos φ k ⃗ \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} r θ × r φ = ∣ ∣ i − 4 sin φ sin θ 4 cos φ cos θ j 4 sin φ cos θ 4 cos φ sin θ k 0 − 4 sin φ ∣ ∣ = − 16 sin 2 φ cos θ i − 16 sin φ cos φ sin 2 θ k − 16 sin 2 φ sin θ j − 16 sin φ cos φ cos 2 θ k = − 16 sin 2 φ cos θ i − 16 sin 2 φ sin θ j − 16 sin φ cos φ ( sin 2 θ + cos 2 θ ) k = − 16 sin 2 φ cos θ i − 16 sin 2 φ sin θ j − 16 sin φ cos φ k
Hence the magnitude
∥ r ⃗ θ × r ⃗ φ ∥ = 1 6 2 sin 4 φ cos 2 θ + 1 6 2 sin 4 φ sin 2 θ + 1 6 2 sin 2 φ cos 2 φ = 1 6 2 sin 4 φ ( cos 2 θ + sin 2 θ ) + 1 6 2 sin 2 φ cos 2 φ = 1 6 2 sin 2 φ ( sin 2 φ + cos 2 φ ) = 16 sin φ \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} ∥ r θ × r φ ∥ = 1 6 2 sin 4 φ cos 2 θ + 1 6 2 sin 4 φ sin 2 θ + 1 6 2 sin 2 φ cos 2 φ = 1 6 2 sin 4 φ ( cos 2 θ + sin 2 θ ) + 1 6 2 sin 2 φ cos 2 φ = 1 6 2 sin 2 φ ( sin 2 φ + cos 2 φ ) = 16 sin φ
The surface integral is
∬ S z d S = ∬ D 4 cos φ ( 16 sin φ ) d A = ∫ 0 2 π ∫ 0 π 2 32 sin ( 2 φ ) d φ d θ = ∫ 0 2 π ( − 16 cos ( 2 φ ) ) ∣ 0 π 2 d θ = ∫ 0 2 π 16 d θ = 32 π \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} S ∬ z d S = D ∬ 4 cos φ ( 16 sin φ ) d A = ∫ 0 2 π ∫ 0 2 π 32 sin ( 2 φ ) d φ d θ = ∫ 0 2 π ( − 16 cos ( 2 φ ) ) ∣ ∣ 0 2 π d θ = ∫ 0 2 π 16 d θ = 32 π
#339914 on Dec 2023
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