# Integral field vector

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• Apr 27th 2009, 08:07 AM
Apprentice123
Integral field vector
Explain how to relate the integral curvilinear of field vector $F(x,y) = (x-y)i + (x+y)j$ with the area the region limited by a regular curve, simple and closed ? Apply the result and calculate the area of the ellipse:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
• Apr 27th 2009, 11:05 AM
Calculus26
Consider Green's Theorem

For F = (x-y) i +(x +y) j

d(x+y)/dx = 1 d(x-y)/dy = -1

The line integral then is 2times the double integral over R , or twice the area of the region enclosed by your ellipse

It is easier to compute this with the line integral than with a double integral.

so the area enclosed is 1/2 the line integral

parameterize C with x = acos(t) y = bsin(t)

and now you are good to go.

Remember you should get pi*ab
• Apr 27th 2009, 11:59 AM
Apprentice123
Quote:

Originally Posted by Calculus26
Consider Green's Theorem

For F = (x-y) i +(x +y) j

d(x+y)/dx = 1 d(x-y)/dy = -1

The line integral then is 2times the double integral over R , or twice the area of the region enclosed by your ellipse

It is easier to compute this with the line integral than with a double integral.

so the area enclosed is 1/2 the line integral

parameterize C with x = acos(t) y = bsin(t)

and now you are good to go.

Remember you should get pi*ab

I find $2 \pi ab$ because my integral $\int_0^{2 \pi}$ the correct is $\int_0^{ \pi}$ ?
• Apr 27th 2009, 12:06 PM
Calculus26
No-- remember the area is 1/2 the line integral--so 0 t0 2pi is correct

if you use 0 to pi you don't have a closed curve
• Apr 27th 2009, 12:10 PM
Apprentice123
Quote:

Originally Posted by Calculus26
No-- remember the area is 1/2 the line integral--so 0 t0 2pi is correct

if you use 0 to pi you don't have a closed curve

Ok thank you the answer is $2 \pi ab$
• Apr 27th 2009, 12:13 PM
Calculus26
No the answer is pi *ab The line integral is 2pi*ab and the AREA is 1/2 this
• Apr 27th 2009, 12:22 PM
Apprentice123
Quote:

Originally Posted by Calculus26
No the answer is pi *ab The line integral is 2pi*ab and the AREA is 1/2 this

why the AREA is $\frac{1}{2}$ ?
• Apr 27th 2009, 12:25 PM
Calculus26
See my first post on this -- I explained it there
• Apr 27th 2009, 12:28 PM
Apprentice123
Quote:

Originally Posted by Calculus26
See my first post on this -- I explained it there

Ok te green theorem is 2 AREA 1/2. If green theorem is 1 AREA 1, if 3 AREA 1/3 ??
• Apr 27th 2009, 12:31 PM
Calculus26
That would make sense wouldn't it ?

If F = f i +g j

And if dg/dx- df/dy is a constant k then Green's theorem gives k *Area
• Apr 27th 2009, 12:38 PM
Apprentice123
Quote:

Originally Posted by Calculus26
That would make sense wouldn't it ?

If F = f i +g j

And if dg/dx- df/dy is a constant k then Green's theorem gives k *Area

Yes thank you