- #1
ninevolt
- 21
- 0
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:
[itex]\int \int \frac{R}{h}z \sqrt{\frac{R^{2}}{h^{2}} + 1} dz d\theta[/itex]
Where:
R = radius of the base
h = height of the cone
(R/h)z = radius of cone at specific z
[itex]\sqrt{\frac{R^{2}}{h^{2}} + 1}[/itex] - the ds element across the slanted side of the cone
and I will obtain the correct answer for the surface area of a cone:
[itex]\pi R \sqrt{h^{2} + R^{2}}[/itex]
but when I try to do the same integral in spherical coordinates I obtain different results
I use the following integral:
[itex]\int \int \rho^{2} sin(\theta) d\rho d\phi [/itex]
What am I doing wrong?
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:
[itex]\int \int \frac{R}{h}z \sqrt{\frac{R^{2}}{h^{2}} + 1} dz d\theta[/itex]
Where:
R = radius of the base
h = height of the cone
(R/h)z = radius of cone at specific z
[itex]\sqrt{\frac{R^{2}}{h^{2}} + 1}[/itex] - the ds element across the slanted side of the cone
and I will obtain the correct answer for the surface area of a cone:
[itex]\pi R \sqrt{h^{2} + R^{2}}[/itex]
but when I try to do the same integral in spherical coordinates I obtain different results
I use the following integral:
[itex]\int \int \rho^{2} sin(\theta) d\rho d\phi [/itex]
What am I doing wrong?
