# Vector Calculus finding the Surface element of a shape

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

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: $$dS = r \csc \phi\, dr\, d\theta$$