Problems from Calculation of surface areas in space

Development:

First of all, we need the parametric form. We use the following parametrization of the sphere: $r(\phi ,\theta )=(\sin \phi \cos \theta ,\sin \phi \cos \theta ,\cos \phi )$.

The statement says that the region is limited by two meridians, ${\theta }_1$ and ${\theta }_2$. Therefore, the interval of $\theta$ is $\theta \in [{\theta }_1,{\theta }_2]$.

We also know that the region is limited by the parallels $z=0$ and $z={\displaystyle \frac{1}{2}}$. Namely, $0⩽\cos \phi ⩽{\displaystyle \frac{1}{2}}\Rightarrow {\displaystyle \frac{\pi }{3}}⩽\phi ⩽{\displaystyle \frac{\pi }{2}}$.

Once we have the parametric form for the region, let's calculate the derivative and the module of its vector product:

$\begin{array}{l}{r}_{\phi }=(\cos \theta \cos \phi ,\sin \theta \cos \phi ,-\sin \phi )\\ r_{\theta }=(-\sin \theta \sin \phi ,\sin \phi \cos \theta ,0)\end{array}\}$

$\begin{array}{lcl}\Rightarrow r_{\phi }\times r_{\theta }& =& \left|\begin{array}{ccc}i& j& k\\ \cos \theta \cos \phi & \sin \theta \cos \phi & -\sin \phi \\ -\sin \theta \sin \phi & \sin \phi \cos \theta & 0\end{array}\right|\\ & =& ({\sin }^2\phi \cos \theta ,{\sin }^2\phi \sin \theta ,\sin \phi \cos \theta )\\ \Rightarrow |r_{\phi }\times r_{\theta }|& =& \sqrt{{\sin }^4\phi {\sin }^2\theta +{\sin }^4\phi {\cos }^2\phi +{\sin }^2\phi {\cos }^2\theta }\\ & =& \sqrt{{\sin }^4\phi +{\sin }^2\phi {\cos }^2\phi }=\sin \phi \end{array}$

Finally, let's integrate:

$$\begin{array}{rl}\text{Area}(S)=& {\int }_{S}dS=\iint |r_{\phi }\times r_{\theta }|d\phi d\theta ={\int }_{{\theta }_1}^{{\theta }_2}{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\sin \phi d\phi d\theta \\ =& {\int }_{{\theta }_1}^{{\theta }_2}[-\cos \phi {]}_{\frac{\pi }{3}}^{\frac{\pi }{2}}d\theta ={\displaystyle \frac{1}{2}}({\theta }_2-{\theta }_1)={\displaystyle \frac{\pi }{4}}\end{array}$$

Solution:

$\text{Area}(S)={\displaystyle \frac{\pi }{4}}$

Hide solution and development