Given the sphere $x^2 + y^2 + z^2 = 1$ and the cone $z = \alpha \sqrt{x^2 + y^2}$, $\alpha > 0$, $S_1$ is the portion of the sphere inside the cone while $S_2$ is the portion of the cone inside the sphere. I need to find the value of $\alpha$ such that both surface areas are the same (it should be $\alpha = 3/4$).

This confuses me not because of the parametrizations by themselves, but because of the limits of integration and the method of simplyfying $\alpha$ for a numerical value.

For $S_{1}$, we can parametrize the sphere and define the domain as we see fit in order to restrict for the cone. Then, $r_1(u,v) = \left(cos(u)sin(v),sin(u)sin(v),cos(v) \right)$. $S_2$ is similar: $r_2(u,v) =\left(u cos(v), u sin(v), \alpha u \right) $, where z is obtained by plugging in the radius in the given cone's equation.

This is where things get confusing. There are two ways to obtain the element of area: $\mid{r_u * r_v}\mid$ and the one I've had more sucess with in this particular problem: $z = f(x,y)$ -> $\sqrt{1 + z_x^2 + z_y}$. This makes us go back to the initial equations ignoring the parametrizations for now. Also, since the cone is in the positive direction, our other z is simply $z = \sqrt{1 - x ^2 - y^2}$. Thus,

$S_{1}$: $\sqrt{1 + \dfrac{x^2}{1-x^2 -y^2 } + \dfrac{y^2}{1-x^2 -y^2} } = \sqrt{\dfrac{1}{1-x^2 -y^2 }} $

$S_{2}$: $\sqrt{1 + \dfrac{a^2 x^2}{x^2 + y^2} + \dfrac{a^2 y^2}{x^2 + y^2} } = \sqrt{1 + a^2} $

I don't know what to do next. According to the brief solution I have at hand, the limit of integration for both $S_1$ and $S_2$ is $x^2 + y^2 \leq a^2$, but I don't know where that comes from (although a quick sketch of the zone where the sphere and the cone touch shows that the intersection would be indeed a circle) and, even more confusing, is a step in the second integral that I can't understand:

$\int \int_{x^2 + y^2 \leq a^2 } \sqrt{1 +a^2} dxdy = \pi a^2 \sqrt{1+ a^2} = \dfrac{\pi}{\sqrt{1 + a^2}} (?) $ (Where did the last one come from?)

This is more algebra than anything else, but that simplification and one that comes after that (for finding a) seems to assume that $a^2 = \dfrac{1}{1+a^2}$ which, to me, doesn't make any sense. I've spent hours trying to a) understand why the $a^2$ simplification is true and b) the limits of integration without any success whatsoever, which is why I'm posting this exercise that may look somewhat trivial at first glance, but isn't to me.

  • $\begingroup$ How do we sketch this? $S_1$ consists of a frustum of the cone and $S_2$ are two polar caps. Right? $\endgroup$
    – Narasimham
    Oct 24, 2018 at 3:53

3 Answers 3


convert to spherical

$x = \rho\cos\theta\sin \phi\\ y = \rho\sin\theta\sin \phi\\ z = \rho\cos \phi$

$z= \alpha \sqrt{x^2 + y^2}$ becomes

$\cos \phi = \alpha \sin \phi\\ \tan\phi = \frac {1}{\alpha}\\ \phi = \arctan \frac 1{\alpha}$

$\int_0^{2\pi}\int_0^{\arctan{\frac 1\alpha}} \sin\phi \ d\phi\ d\theta$

Is the area of the spherical cap.

$\int_0^{2\pi}\int_0^{1} \rho\sin(\arctan{\frac 1\alpha}) \ d\rho\ d\theta$

Is the area of the cone inside the sphere.

$\pi \sin(\arctan{\frac 1\alpha}) = 2\pi(1-\cos\arctan{\frac 1\alpha})$

$\frac {1}{\sqrt{1+\alpha^2}} = 2(1-\frac {a}{\sqrt {1+\alpha^2}})\\ 1 + 2\alpha = 2\sqrt {1+\alpha^2}\\ 1 + 4\alpha + 4\alpha^2 = 4 +4\alpha^2\\ 4\alpha - 3= 0\\ \alpha = \frac 34$


Assume that a point $(r,0,r\alpha)$ is in the intersection. Then $$ {\rm area}\ S_1=2\pi (1-r\alpha ) $$

And ${\rm area}\ S_2 =\pi \frac{ {\rm length}\ S_2 }{2\pi } = \pi r$ so that $2=r(1+2\alpha )$.


Draw a sketch of cone and polar cap. Cone vertex at sphere center. Semi vertical angle of cone is $ \beta, \cot \beta = \alpha $

Cone lateral area is $\pi R L,$ sphere area is $ 2\pi R h$, $h$ is the polar cap height.

$$ S_1= 2 \pi R^2 (1-\cos \beta),\, S_2 = 2 \pi R^2 \sin \beta\tag1$$

Equate and simplify using half angles $$ \tan (\beta/2)= \frac12 = t. \tag2 $$

$$ \cot \beta = \frac {1-t^2}{2 t } =\frac {4}{3 } = \frac {1}{\alpha} \tag3 $$


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