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Daniel Nierro
science forum beginner
Joined: 19 Jul 2006
Posts: 2
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 Posted: Thu Jul 20, 2006 2:47 pm Post subject:
Curve integral - correct or not?
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Hi!
If one would like to calculate the curve integral of the function f(x,y,z) =
x^2 + y^2 over the curve C: r(t) = (e^t cos(t), e^t sin(t), t) where t goes
from 0 to 1, what would the result be?
The curve is clearly somewhat spiral-shaped with a radius increasing with t,
and the problem should be easily solvable using cylindrical coordinates.
I'm wondering, does e^(2t) sqrt(e^(2t) + 1) sound like a reasonable answer?
Cheers,
Doug  |
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G.E. Ivey
science forum Guru
Joined: 29 Apr 2005
Posts: 308
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| Posted: Thu Jul 20, 2006 6:54 pm Post subject:
Re: Curve integral - correct or not?
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| Quote: | Hi!
If one would like to calculate the curve integral of
the function f(x,y,z) =
x^2 + y^2 over the curve C: r(t) = (e^t cos(t), e^t
sin(t), t) where t goes
from 0 to 1, what would the result be?
The curve is clearly somewhat spiral-shaped with a
radius increasing with t,
and the problem should be easily solvable using
cylindrical coordinates.
I'm wondering, does e^(2t) sqrt(e^(2t) + 1) sound
like a reasonable answer?
Cheers,
Doug :-)
The curve is given by x= e^t cos(t), y= e^t sin(t), and z= t so the integrand, x^2+ y^2= (e^t cos(t))^2+ (e^t sin(t))= e^(2t)(cos^2(t)+ sin^2(t))= e^2t. |
The integral, then, is integral, from t= 0 to 1 e^(2t)dt.
I can see no reason for a sqrt in the answer- and the integral from t= 0 to 1 is a number, not a function of t. |
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Narcoleptic Insomniac
science forum Guru
Joined: 02 May 2005
Posts: 323
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| Posted: Fri Jul 21, 2006 1:46 am Post subject:
Re: Curve integral - correct or not?
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On Jul 20, 2006 9:47 AM CT, David wrote:
| Quote: | Hi!
If one would like to calculate the curve integral of
the function f(x,y,z) = x^2 + y^2 over the curve
C: r(t) = (e^t cos(t), e^t sin(t), t) where t goes
from 0 to 1, what would the result be? The curve is
clearly somewhat spiral-shaped with a radius increasing
with t, and the problem should be easily solvable using
cylindrical coordinates. I'm wondering, does
e^(2t) sqrt(e^(2t) + 1) sound like a reasonable answer?
Cheers,
Doug
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Okay, so we're given that f(x, y, z) = x^2 + y^2 and
C: r(t) = [e^t cos(t), e^t sin(t), t] for 0 <= t <= 1
...and we need to compute int_C f ds. Taking the
derivatives of each component of r yeilds
x'(t) = e^t cos(t) - e^t sin(t),
y'(t) = e^t sin(t) + e^t cos(t), and
z'(t) = 1.
Now by definition ds^2 = dx^2 + dy^2 + dz^2 so
ds^2 = (e^t cos(t) - e^t sin(t))^2
+ (e^t cos(t) + e^t sin(t))^2 + 1
= e^(2t) * (cos(t)^2 - 2cos(t)sin(t) + sin(t)^2)
+ e^(2t) * (cos(t)^2 + 2cos(t)sin(t) + sin(t)^2) + 1
= 2e^(2t) + 1
...which implies...
int_C f ds = int_0^1 e^(2t) * [2e^(2t) + 1]^(1/2) dt.
If we let u = 2e^(2t) + 1, then du = 4e^(2t) dt so
int_0^1 e^(2t) * [2e^(2t) + 1]^(1/2) dt =
1/4 * int_3^{2e^2 + 1} u^(1/2) du =
1/4 * 2/3 u^(3/2) |_3^{2e^2 + 1} =
1/6 * [(2e^2 + 1)^(3/2) - 3^(3/2)].
Regards,
Kyle Czarnecki |
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