Scalar Surface Integral over parameterized surface

  • #1
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Homework Statement



Calculate ∫∫ f(x,y,z)dS for the surface G(r,θ) = (rcosθ,rsinθ,θ), 0<r<1, 0<θ<2pi.
f(x,y,z) = (x^2+y^2)^(1/2) = r

Homework Equations





The Attempt at a Solution


so the surface is given so I have to find the normal vector...

G_r = cos(θ),sinθ,0
G_θ = -rsin(θ),rcos(θ),1

G_r = G_θ = normal vector = sin(θ)I - cos(θ)j + rk

magnitude of normal vector
=
(1+r^2)^(1/2)

heres my main question, usually if I parameterize in terms of u and v this would be like

dS = |Normal vector| du dv

but since this is in terms of r and θ do I just do

dS = |Normal vector| dr dθ
or
dS = |Normal Vector| r dr dθ

cause earlier in this course when we went from rectangular coordinates to polar we had to add an r, does that still apply here?

So basically would my final integral be

∫∫r(1+r^2)drdθ
or
∫∫r^2(1+r^2)drdθ

thanks PF :D
 

Answers and Replies

  • #2
tiny-tim
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Hi PsychonautQQ! :smile:
… usually if I parameterize in terms of u and v this would be like

dS = |Normal vector| du dv

but since this is in terms of r and θ do I just do

dS = |Normal vector| dr dθ
or
dS = |Normal Vector| r dr dθ

cause earlier in this course when we went from rectangular coordinates to polar we had to add an r, does that still apply here?
i] from dimensions, it obviously can't just be dr dθ: dS is length-squared, and dr dθ isn't, so you need at least another r :wink: (though that doesn't tell you whether there's also a function of θ)

ii] more rigorously: you need to find the area of a rectangle (if the coordinates are orthogonal, otherwise it's a more general parallelogram) with sides du and dv

if S was a disc, it would be rdr dθ … is that still true for this actual twisty S?
 
  • #3
LCKurtz
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Homework Statement



Calculate ∫∫ f(x,y,z)dS for the surface G(r,θ) = (rcosθ,rsinθ,θ), 0<r<1, 0<θ<2pi.
f(x,y,z) = (x^2+y^2)^(1/2) = r

Homework Equations





The Attempt at a Solution


so the surface is given so I have to find the normal vector...

G_r = cos(θ),sinθ,0
G_θ = -rsin(θ),rcos(θ),1

G_r = G_θ = normal vector = sin(θ)I - cos(θ)j + rk

magnitude of normal vector
=
(1+r^2)^(1/2)

heres my main question, usually if I parameterize in terms of u and v this would be like

dS = |Normal vector| du dv

but since this is in terms of r and θ do I just do

dS = |Normal vector| dr dθ
or
dS = |Normal Vector| r dr dθ
You use the first one. No extra ##r##
[Edit, added]: So that would be ##dS=\sqrt{1+r^2}drd\theta##. Neither of the two below are correct. I didn't notice that the first time. Then you get an extra ##r## for the function ##f##.

cause earlier in this course when we went from rectangular coordinates to polar we had to add an r, does that still apply here?

So basically would my final integral be

∫∫r(1+r^2)drdθ
or
∫∫r^2(1+r^2)drdθ

thanks PF :D
What happens often when teaching calculus is that polar coordinates and areas are taught before the parameterization stuff. So you memorize the ##rdrd\theta## as the polar element of area and your text or teacher may give an intuitive argument for why the ##r## is in there. In fact, the ##r## comes from the polar coordinate transformation in the ##xy## plane:$$
\vec R(r,\theta) = \langle r\cos\theta,r\sin\theta,0\rangle$$If you calculate ##|\vec R_r\times\vec R_\theta|## for this parameterization as you did above, you will get ##r## so ##dS = rdrd\theta## is the polar surface element in the ##xy## plane. So the upshot is if you use ##dS =|\vec R_u\times \vec R_v|dudv## you never add anything else, such as an extra ##r## in this case.
 
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  • #4
LCKurtz
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Hi PsychonautQQ! :smile:


i] from dimensions, it obviously can't just be dr dθ: dS is length-squared, and dr dθ isn't, so you need at least another r :wink: (though that doesn't tell you whether there's also a function of θ)
In his example, the additional ##r## is built into the ##|r_r \times r_\theta|## so his first formula is correct as it stands.
 
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  • #5
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Wow thanks for the great explanation! Just filled a big gap in my conceptual understanding.
so my final integral will look like
∫∫r^2(1+r^2)drdθ with dr between 0 and 1 and theta between 0 and 2pi? Sorry i'm being so picky about making sure it's right i'm also part of a group and I want to be useful when we meet up tomorrow
 
  • #6
LCKurtz
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Wow thanks for the great explanation! Just filled a big gap in my conceptual understanding.
so my final integral will look like
∫∫r^2(1+r^2)drdθ with dr between 0 and 1 and theta between 0 and 2pi? Sorry i'm being so picky about making sure it's right i'm also part of a group and I want to be useful when we meet up tomorrow
Please note the edited correction in post #3.
 
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  • #7
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oh right! of course. okay I made that correction so it's now
∫∫r^2(1+r^2)^(1/2)drdθ dr from 0 to 1 and d(theta) from 2 to 2pi. Thanks yo! (not an easy integral to evaluate though, and i'm not suppose to use the internet X_x)
 
  • #8
LCKurtz
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It still isn't right. Your ##f(x,y,z)= r##. That's all that gets multiplied by the ##dS##, which has no ##r## factor.
 
  • #9
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riiight... golly >.<

∫∫r(1+r^2)^(1/2)drdθ

where dS = (1+r^2)^(1/2) drdθ
 
  • #10
LCKurtz
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Right. And the bonus is that it's now easy to integrate.
 
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  • #11
tiny-tim
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In his example, the additional ##r## is built into the ##|r_r \times r_\theta|## so his first formula is correct as it stands.
You use the first one. No extra ##r##
[Edit, added]: So that would be ##dS=\sqrt{1+r^2}drd\theta##.
yes of course … i somehow thought that the "normal vector" was the unit normal vector, which of course it wasn't :redface:

thanks :smile:
 
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