What is the surface integral of a cone in polar coordinates?

[XcKyle93]

Homework Statement

g(x,y,z) = z²; Ʃ is the part of the cone z = [itex]\sqrt{x²+y²}[/itex] between the planes z = 1 and z = 3.

Homework Equations

Conversion to polar coordinates
∫∫_Ʃg(x,y,z)dS = ∫∫_Rg(x,y,f(x,y)) [itex]\sqrt{f_x² + f_y²+1}[/itex]

The Attempt at a Solution

If we're talking in terms of r and θ, r goes from 1 to 3 and θ goes from 0 to 2π. I converted to polar coordinates from cartesian and I got 8*π*√2 *z². If you were converting to polar/cylindrical coordinates, z² wouldn't change, correct?

Sorry, for some reason the square root LaTeX command didn't want to work...

 

[LCKurtz]

Homework Statement

g(x,y,z) = z²; Ʃ is the part of the cone z = $\sqrt{x^2+y^2}$ between the planes z = 1 and z = 3.

Homework Equations

Conversion to polar coordinates
∫∫_Ʃg(x,y,z)dS = ∫∫_Rg(x,y,f(x,y)) $\sqrt{f^2_x + f^2_y+1}$

The Attempt at a Solution

If we're talking in terms of r and θ, r goes from 1 to 3 and θ goes from 0 to 2π. I converted to polar coordinates from cartesian and I got 8*π*√2 *z². If you were converting to polar/cylindrical coordinates, z² wouldn't change, correct?

Sorry, for some reason the square root LaTeX command didn't want to work...

Don't use the subscript and superscript buttons in a tex expression use f_x for subscripts and f^2 for superscripts in tex as I have done for you.

You shouldn't have any z in your answer; it should be a number. Of course z depends on x and y or r and θ because on the surface $z=\sqrt{x^2+y^2}$.