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Narcoleptic Insomniac
science forum Guru

Joined: 02 May 2005
Posts: 323 Posted: Fri Jul 21, 2006 1:46 am    Post subject: Re: Curve integral - correct or not? 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 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 G.E. Ivey
science forum Guru

Joined: 29 Apr 2005
Posts: 308 Posted: Thu Jul 20, 2006 6:54 pm    Post subject: Re: Curve integral - correct or not? 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. Daniel Nierro
science forum beginner

Joined: 19 Jul 2006
Posts: 2 Posted: Thu Jul 20, 2006 2:47 pm    Post subject: Curve integral - correct or not? 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   Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
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