integration
I have successfully found the surface area of a cone (without the base) using integration and cylindrical coordinates.
Next, I was asked to find it using spherical coordinates and the following parameters:
$\theta$ – half of the constant opening angle of the cone.
$h$ – the height of the cone.
$R=h\tan(\theta)$ – the radius of the base.
I've tried the following: $$\int_{0}^{R}2\pi cos(\theta)rdr $$
But obviously, it produces a wrong answer.
Any ideas?
A picture always helps. The green region is the one you are interested in (assuming sphere centered at origin)
The limits of integration for spherical coordinates as you rightly noted in comments in the other post are $0\leq r \leq 6$ and $0\leq \theta \leq 2\pi$
For finding the limits for $\phi$, refer to the picture and note that the problem specified that the angle of the black triangle at the vertex is $90^\circ = \pi/2$. Consider how much of an angle from the red line you will need to go down (hint: the red line bisects the angle) to find the largest $\phi$ can be.
Your answer gives you the volume of a cylinder with height $2$.
See the picture:
$$\frac{2}{\rho}=\cos \phi\implies \rho=\frac{2}{\cos\phi}=2\sec\phi$$
Best Answer
Your problem gets a bit confusing because you're using $\theta$ to represent what's usually $\phi$ in spherical coordinates. Where are you putting the apex of your cone? Imagine an incredibly tall and thin cone: would the limits of your integration with respect to $r$ still be $0$ and the radius $R$, even though the slant height $\sqrt{R^2+h^2}$ is much greater than $R$?
HINT: if we represent the spherical coordinates as $(r,\alpha,\beta)$, where $(x,y,z)$ in Cartesian coordinates is $(r \sin \beta \cos \alpha, r \sin \beta \sin \alpha, r \cos \beta)$, and if you put the apex of the cone at the origin facing upward, then the entire surface area of the cone has a constant $\beta$-coordinate $\beta = \theta$.
Since $\beta$ is constant, all we care about are $r$ and $\alpha$, and small changes in $r$ and $\alpha$ just draw out small slanted trapezoids on the surface of the cone, whose width (around the cone) is $r\: d\alpha$ and whose height (down the cone) is $\sin \theta \: dr$ from basic trigonometry (try drawing a cross-section of the cone and see what $\sin \theta \: dr$ represents). Now find the limits of integration for $\alpha$ and $r$, find the value of the integral, find $\sin \theta$ in terms of $R$ and $h$, and you'll find yourself with the answer!