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: $$ r(\varphi,\theta)=(\sin\varphi\cos\theta, \sin\varphi\cos\theta, \cos\varphi)$$.

The statement says that the region is limited by two meridians, $$\theta_1$$ and $$\theta_2$$. Therefore, the interval of $$\theta$$ is $$\theta\in\big[ \theta_1,\theta_2 \big]$$.

We also know that the region is limited by the parallels $$z = 0$$ and $$z = \dfrac{1}{2}$$. Namely, $$0\leqslant\cos\varphi \leqslant \dfrac{1}{2} \Rightarrow \dfrac{\pi}{3}\leqslant\varphi\leqslant\dfrac{\pi}{2}$$.

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

$$ \left. \begin{array}{l} r_\varphi=\big(\cos\theta\cos\varphi, \sin\theta\cos\varphi,-\sin\varphi\big) \\ r_\theta=\big(-\sin\theta\sin\varphi, \sin\varphi\cos\theta,0\big) \end{array} \right\} $$

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

Finally, let's integrate:

$$$ \begin{array}{rl} \text{Area}(S)=&\int_S dS=\iint |r_\varphi \times r_\theta| \ d\varphi \ d\theta= \int_{\theta_1}^{\theta_2}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sin\varphi \ d\varphi \ d\theta \\ =& \int_{\theta_1}^{\theta_2}[-\cos\varphi]_{\frac{\pi}{3}}^{\frac{\pi}{2}} \ d\theta = \dfrac{1}{2} (\theta_2-\theta_1)= \dfrac{\pi}{4} \end{array} $$$

Solution:

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

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