I have to
Compute the surface area of that portion of the sphere $x^2+y^2+z^2=a^2$ lying within the cylinder $\Bbb{T}:=\ \ x^2+y^2=by.$
My work:
I start with only the $\Bbb{S}:=\ \ z=\sqrt{a^2-x^2-y^2}$ part and will later multiply it by $2$.
$${\delta z\over \delta x}={-x\over \sqrt{a^2-x^2-y^2}}\ ;\ {\delta z\over \delta y}={-y\over \sqrt{a^2-x^2-y^2}}$$
Using the formula $$a(\Bbb{S})=\iint\limits_\Bbb{T}\sqrt{1+\left({\delta z\over \delta x}\right)^2+\left({\delta z\over \delta y}\right)^2}dx \ dy$$
I get $$a(\Bbb{S})=\iint\limits_{x^2+y^2=by}\sqrt{1+{by\over a^2-by}}\ dy\ dx\\=\int_0^{b/2}\int\limits_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}\sqrt{1+{by\over a^2-by}}\ dy\ dx\\=a\int_0^{b/2}\int\limits_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}{1\over\sqrt{a^2-by}}dy \ dx\\=a\int_0^{b/2}\left\{\left[{-2\sqrt{a^2-by}\over b}\right]_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}\right\}\ dx$$
How to proceed now?The integral seems too bad.
OR
Is there a simpler parametrization ?
Best Answer
Another approach in spherical coordinates: parametrize the surface with \begin{cases} x=a \sin\phi \cos\theta \\ y=a \sin\phi \sin\theta \\ z=a \cos\phi \end{cases}
with $(\theta,\phi)\in [0,\pi]\times[0,\Phi]$, where $\Phi \in ]0,\pi]$ is the solution to \begin{cases} x^2+y^2+z^2=a^2 \\ x^2+y^2=by \end{cases} Substituting $x,y,z$ by their expressions in spherical coordinates yields $$ \Phi = \sin^{-1}\left(\frac{b}{a}\sin\theta\right) $$ It follows that $$ S=\int_{0}^{\pi}\int_0^{\Phi} a^2\sin\phi\; d\phi d\theta = a^2 \int_0^{\pi}1-\cos\Phi\; d\theta $$ With a little trigonometric work ($\cos x =\pm \sqrt{1-\sin^2x}$), we can show that $$ \cos\Phi = \sqrt{1-\left(\frac{b}{a}\sin\theta\right)^2} $$ Therefore $$ S = \pi a^2 - a \int_0^{\pi} \sqrt{a^2-b^2\sin^2\theta}\; d\theta $$