How to calculate area of curved surface in specific region

Question

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Answer 1

We'll work in spherical coordinates. Firstly, convert the given Cartesian equations to spherical:

Sphere:

$\quad \rho = a$

Cylinder:

Note that the cylinder has radius $\frac{a}{2}$ with centre at $(x,\;y) = (\frac{a}{2},\;0)$. Angle $\theta$ ranges from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, and $\phi$ ranges from $0$ to $\pi$.

For any point on the cylinder, $\rho = \dfrac{\sqrt{x^2+y^2}}{\sin{\phi}} = \dfrac{\sqrt{ax}}{\sin{\phi}}$. Now, $\cos^2{\theta} = \dfrac{x^2}{x^2+y^2} = \dfrac{x^2}{ax} = \dfrac{x}{a}$. Therefore, $\sqrt{ax} = a\cos{\theta}$, so the equation of the cylinder is,

$$\rho = \dfrac{a\cos{\theta}}{\sin{\phi}}.$$

$$\\$$

For the intersection of the sphere and cyinder, combining the above two equations gives,

\begin{eqnarray*} a &=& \dfrac{a\cos{\theta}}{\sin{\phi}} \\ \therefore \cos{\theta} &=& \sin{\phi} = \cos{(\phi - \frac{\pi}{2})} \\ \therefore \theta &=& \pm (\phi - \frac{\pi}{2}). \end{eqnarray*}

This means the required surface has $\theta$ bounded above and below by the two values $\pm (\phi - \frac{\pi}{2})$. To find the area, let $\;dA = a^2\sin{\phi} \;d\theta \;d\phi\;$ be the surface area of a rectangular wedge originating from the origin, in the expected way. Hopefully, you know this formula or how to derive it. Note that the required surface is symmetrical about both the $xy-\text{plane}$ and the $xz-\text{plane}$. So we can calculate the area of one of the quarters of it and multiply by $4$. This explains the $0$ as the lower limit of integration for $\theta$ and $\frac{\pi}{2}$ as the upper limit for $\phi$ in the integral below. The area required then is,

\begin{eqnarray*} A &=& 4 \int_{\phi = 0}^{\frac{\pi}{2}} \int_{\theta = 0}^{\frac{\pi}{2} - \phi} {a^2 \sin{\phi} \;d\theta \;d\phi} \\ && \\ &=& 4 a^2 \int_{0}^{\frac{\pi}{2}} {\left[ \;\theta \;\right]_{0}^{\frac{\pi}{2} - \phi} \sin{\phi} \;d\phi} \\ && \\ &=& 4 a^2 \int_{0}^{\frac{\pi}{2}} {\left( \frac{\pi}{2} - \phi \right) \sin{\phi} \;d\phi} \\ && \\ &=& 4 a^2 \left[ \phi \cos{\phi} - \sin{\phi} - \frac{\pi}{2}\cos{\phi} \right]_{0}^{\frac{\pi}{2}} \\ && \\ &=& 4 a^2 \left((0 - 1 - 0) - (0 - 0 - \frac{\pi}{2}) \right) \\ && \\ &=& 2 a^2 \left(\pi - 2 \right).\end{eqnarray*}