Circulation

**Apprentice123** · June 8th 2009, 01:34 PM

Since the field vectors $F(x,y,z) = (x-y)i + (y-z)j + (z-x)k$ and the curve $\alpha$ that is the intersection of surfaces:
$S1: x+y+z =1$ and $S2: x^2+y^2=1$
Find the circulation of F arrond the $\alpha$

My solution:

curl F = (1,1,1)

Surface S1:

Normal: $N = (0,0,1)$

$\int_0^1 \int_0^{1-v} dudv = \frac{1}{2}$

Surface S2:

Normal: $N = (cosu, sinu, 0)$

$\int_0^{2 \pi} \int_0^1 (rcos \theta + rsin \theta)rdrd \theta = 0$

Circulation:

$\int_C F.dr = \int \int_S F.N.da = S1 + S2 = \frac{1}{2}$

It is correct ?

**Calculus26** · June 9th 2009, 08:12 AM

No-- you kind of missed the point

alpha is the curve of intersection of the plane and the cylinder

See the attachment for the details where the line integral is computed directly

If you use Stokes theorem then your surface is z= 1-x-y over a circle of radius 1

N = i + j + k

curl F*N = 3 int(curlF*NdA) = 3*area = 3 pi which is what we obtain computing the line integral directly

**Apprentice123** · June 9th 2009, 11:19 AM

Originally Posted by Calculus26

No-- you kind of missed the point

alpha is the curve of intersection of the plane and the cylinder

See the attachment for the details where the line integral is computed directly

If you use Stokes theorem then your surface is z= 1-x-y over a circle of radius 1

N = i + j + k

curl F*N = 3 int(curlF*NdA) = 3*area = 3 pi which is what we obtain computing the line integral directly

Ok. Thank you

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