Area Under Hyperbola: Green's Theorem

In summary, the conversation discusses the use of Green's theorem to find the area enclosed by a hyperbola and a line. The integrand for the linear part of the boundary is different from that of the hyperbola part, and it is recommended to do two separate line integrals instead of trying to combine them. The conversation also mentions the need for a consistent orientation for the two line integrals.
  • #1
sara_87
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Homework Statement



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

Homework Equations



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

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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  • #2
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.
 
  • #3
thanks for the limits, i agree.
When i parametrize the linear part at the boundary x=3, how does this effect the integrand?
 
  • #4
sara_87 said:
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.
 
  • #5
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)
?
 
  • #6
sara_87 said:
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.
 

FAQ: Area Under Hyperbola: Green's Theorem

1. What is the formula for calculating the area under a hyperbola using Green's Theorem?

The formula for calculating the area under a hyperbola using Green's Theorem is A = ∮(Pdx + Qdy), where P and Q are the partial derivatives of the hyperbola equation.

2. How does Green's Theorem relate to the area under a hyperbola?

Green's Theorem is a mathematical theorem that relates the area under a curve to the line integral of a two-dimensional vector field over a closed curve. For a hyperbola, the vector field is defined by its partial derivatives, and the line integral represents the area under the curve.

3. Can Green's Theorem be used to find the area under any type of curve?

Yes, Green's Theorem can be used to find the area under any type of curve as long as the curve is closed and the vector field is defined by its partial derivatives. This includes curves such as ellipses, parabolas, and circles.

4. What are the limitations of using Green's Theorem to find the area under a hyperbola?

The limitations of using Green's Theorem to find the area under a hyperbola include the fact that the hyperbola must be closed and the vector field must be defined by its partial derivatives. Additionally, the hyperbola must be oriented in the counterclockwise direction for the theorem to be applicable.

5. How is Green's Theorem different from other methods for finding the area under a hyperbola?

Green's Theorem is different from other methods for finding the area under a hyperbola because it relies on the concept of line integrals and vector fields, rather than traditional methods such as integration by parts. It also allows for the calculation of areas under more complex curves by considering the vector field over the entire curve rather than just the curve itself.

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