Let $S$ be the sphere given by the equation $x^2+y^2 +z^2 =4$ cut with $z \geq 0$. Now, we drill the semisphere that is left with two vertical cylinders of radius $1$, whose axes are respectively on the points $(0,1,0)$ and $(0,-1,0)$. Calculate the area of the surface that is left.

I know that the area of the semisphere of radius two is $8\pi$, but I don't know how to compute the area of the intersection between the cylinders and the semisphere. Any help with that would be highly appreciate.

Thanks in advance!

  • $\begingroup$ @Xoque55 I tried to parametrize the intersection surface as if it was the graph of $f(x,y) = \sqrt { 4 - x^2- y^2}$. But I'm still having some trouble in picturing the surface, so I don't know how to set the domain of $f$. $\endgroup$ – user335485 Jun 11 '16 at 14:45
  • $\begingroup$ @user314159 I updated my answer with full calculation, take a look. $\endgroup$ – Jack's wasted life Jun 29 '16 at 3:51

Parametrization of the surface cut out by one cylinder

enter image description here

The image above shows the intersection of the hemisphere with the cylinder whose axis passes through $(0,1,0)$ and whose equation is $$ x^2+(y-1)^2=1 $$ The surface cut out by the cylinder can be parametrized with Cartesian coordinates as $$ \vec\Sigma=\left(x,y,\sqrt{4-x^2-y^2}\right),\quad 0\le x^2+(y-1)^2\le1\\ \vec\Sigma_x\times\vec\Sigma_y=\begin{vmatrix} \hat i&\hat j&\hat k\\ 1& 0&-{x\over\sqrt{4-x^2-y^2}}\\ 0& 1&-{y\over\sqrt{4-x^2-y^2}}\\ \end{vmatrix}\\ |\vec\Sigma_x\times\vec\Sigma_y|={2\over\sqrt{4-x^2-y^2}}$$

Calculation of surface area cut out by one cylinder

Area of the surface cut out by one cylinder is $$\mathrm{ \int_0^{2}\int^{\sqrt{1-(y-1)^2}}_{-\sqrt{1-(y-1)^2}}{2\over\sqrt{4-x^2-y^2}}\,dx\,dy\\ =2\int_0^{2}\int^{\sqrt{2y-y^2}}_{-\sqrt{2y-y^2}}{1\over\sqrt{4-x^2-y^2}}\,dx\,dy\\ =4\int_0^{2}\int^{\sqrt{2y-y^2}}_{0}{1\over\sqrt{4-y^2-x^2}}\,dx\,dy\\ =4\int_0^2\arcsin{\sqrt{2y-y^2\over4-y^2}}\,dy\\ =4\int_0^2\arctan{\sqrt{2y-y^2\over4-2y}}\,dy\\ =4\int_0^2\arctan\sqrt{y\over2}\,dy\\ =16\int_0^1v\arctan v\,dv\quad\left(v=\sqrt{y\over2}\right)\\ =16\left[{v^2\over2}\arctan v\Big{|}_0^1-\int_0^1{v^2\over2(1+v^2)}dv\right]\\ =16\left[{\pi\over8}-{1\over2}\int_0^1\left(1-{1\over1+v^2}\right)dv\right]\\ =8\left[{\pi\over4}-\left(v-\arctan v\right)\Big{|}_0^1\right]\\=\color{blue}{4(\pi-2)} }$$

Final answer

As $2$ identical holes have been drilled and the area of the hemisphere is ${1\over2}4\pi\times2^2$ sq units, the area of the surface that is left is $$ {1\over2}4\pi\times2^2-2\times\color{blue}{4(\pi-2)}=16 $$


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