Consider a sphere $\Gamma$ centered around the origin and of radius $R$.
Consider a circle $\gamma_0$ of radius $r_0 < R$ on the sphere $\Gamma$. The circle is centered around the $z$-axis and is in a plane parallel to the $(x,y)$ plane, at $z_0 > 0$.
Let $A_0$ be the surface area on the sphere $\Gamma$ bounded by the circle $\gamma_0$ and with $z \ge z_0$. Let $V_0$ be the volume of the ‘ice cream’ cone consisting of all points on line segments joining the origin to points on $A_0$. Let $A_1$ be the surface area of the side of the cone $V_0$.Provide expressions for the area element $dS$ of the ‘ice cream’ cone volume $V_0$ over its surface, $S_0 = A_0 + A_1$.
I can quite easily find the area element corresponding to $A_0$ as this is simply $$r^2\sin\theta\,d\phi \,d\theta \,\bf{r}$$ where $\bf{r}$ is a unit vector. But I am struggling to find out the expression for $A_1$ as the side of the cone doesn't seem logical to use spherical polars to find the surface area.
Any help would be really appreciated. Thanks
Best Answer
I don't know if this is exactly what you're looking for, but if I have a cone whose tip is located at the origin and whose sides slope at an angle $\phi$ away from the positive $z$-axis:
Then you can parametrize this surface using polar coordinates like so: $$ \vec S(r,\theta) = (r \cos \theta,\ r \sin \theta,\ r \cot \phi) $$ The area element is then $$ dS = \left|\frac{d\vec S}{dr} \times \frac{d\vec S}{d\theta} \right| \, dr\, d\theta $$ which yields if you go thru all the trouble: $$ dS = r \csc \phi\, dr\, d\theta $$