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Area under hyperbola

by sara_87
Tags: hyperbola
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sara_87
#1
Nov8-09, 03:59 PM
P: 774
1. The problem statement, all variables and given/known data

Find the area enclosed by the hyperbola: 25x^2-4y^2=100 and the line x=3
using the green's theorem

2. Relevant equations

Green's theorem:
[tex]\int_C[Pdx+Qdy]=\int\int(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy[/tex]

3. The attempt at a solution

We can write the area of the domain as:
area=[tex]\frac{1}{2}\int(xdy-ydx)[/tex]
I know what the graph looks like and i know the parametrisation:
x=2cosht
y=bsinht
but i am to use: area=[tex]\frac{1}{2}\int(xdy-ydx)[/tex] what would be the limits of integration?
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Dick
#2
Nov8-09, 07:18 PM
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The t limits for the hyperbolic segment of the parametrization are where x=3, i.e. 3=2*cosh(t), yes? Don't forget you need a separate parametrization for the linear part of the boundary x=3 and don't forget to choose a consistent orientation for the two line integrals.
sara_87
#3
Nov9-09, 05:02 PM
P: 774
thanks for the limits, i agree.
When i parametrize the linear part at the boundary x=3, how does this effect the integrand?

Dick
#4
Nov9-09, 06:03 PM
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Area under hyperbola

Quote Quote by sara_87 View Post
thanks for the limits, i agree.
When i parametrize the linear part at the boundary x=3, how does this effect the integrand?
The integrand is completely different. To do the line part you need to write an x(t) and y(t) that parametrize the line x=3.
sara_87
#5
Nov9-09, 06:14 PM
P: 774
Oh right i see. so when i do that, when i find x(t) and y(t) for the line, and the X(t) and Y(t) for the hyperbola part, how do i out this in the integrand?
I mean for the xdy part, is this: (x(t)+X(t))dy(t)
?
Dick
#6
Nov9-09, 10:12 PM
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Quote Quote by sara_87 View Post
Oh right i see. so when i do that, when i find x(t) and y(t) for the line, and the X(t) and Y(t) for the hyperbola part, how do i out this in the integrand?
I mean for the xdy part, is this: (x(t)+X(t))dy(t)
?
Why don't you just do two separate line integrals instead of trying to mix them up? That's what I would do.


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