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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?

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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!

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