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Surface integral of a cone

  • Thread starter XcKyle93
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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{fx2 + fy2+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...
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



g(x,y,z) = z2; Ʃ is the part of the cone z = [itex]\sqrt{x^2+y^2}[/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^2_x + f^2_y+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...
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 [itex]z=\sqrt{x^2+y^2}[/itex].
 

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