Finding surface area of cone in spherical coordinates

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The discussion focuses on finding the surface area of a hollow cone using spherical coordinates, where the user successfully calculates it with cylindrical coordinates but encounters issues with spherical coordinates. The integral used in cylindrical coordinates yields the correct surface area formula, πR√(h² + R²). In contrast, the attempt with spherical coordinates results in discrepancies, prompting a question about potential errors in the setup. A participant suggests that the user may be confusing the angle θ with the cone's fixed semi-angle. The conversation also hints at an alternative method of calculating the surface area by slicing and flattening the cone.
ninevolt
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Hello everyone,

I recently tried to find the surface area of a hollow cone (there is no base, like an ice cream cone) using spherical coordinates. With cylindrical coordinates I was able to do this easily using the following integral:

\int \int \frac{R}{h}z \sqrt{\frac{R^{2}}{h^{2}} + 1} dz d\theta
Where:
R = radius of the base
h = height of the cone
(R/h)z = radius of cone at specific z

\sqrt{\frac{R^{2}}{h^{2}} + 1} - the ds element across the slanted side of the cone

and I will obtain the correct answer for the surface area of a cone:
\pi R \sqrt{h^{2} + R^{2}}

but when I try to do the same integral in spherical coordinates I obtain different results
I use the following integral:
\int \int \rho^{2} sin(\theta) d\rho d\phi

What am I doing wrong?
 
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hello ninevolt! :smile:

i think you're confusing θ with the (fixed) semi-angle of the cone :wink:

(btw, you might also like to try doing it without integration, by slicing the cone and flattening it!)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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