Find appropriate parametrization to find area bounded by a curve

In summary, to find the area bounded by the curve x^{2/5}+y^{2/5}=a^{2/5}, we can use the parametrization x=rcos\theta, y=rsin\theta and the corresponding Jacobian |J|=r. This simplifies the double integral and allows us to integrate with respect to r and \theta. The limits of integration can be determined by substituting the parametrization into the original curve equation.
  • #1
wifi
115
1
Problem:

Use an appropraite parametrization [tex]x=f(r,\theta), y=g(r,\theta)[/tex] and the corresponding Jacobian such that [tex] dx \ dy \ =|J| dr \ d\theta[/tex] to find the area bounded by the curve [tex]x^{2/5}+y^{2/5}=a^{2/5}[/tex]

Attempt at a Solution:

I'm not really sure how to find the parametrization. Once I have that, calculating the Jacobian is simple. Then all that's left is computing a double integral. Right?
 
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  • #2
wifi said:
Problem:

Use an appropraite parametrization [tex]x=f(r,\theta), y=g(r,\theta)[/tex] and the corresponding Jacobian such that [tex] dx \ dy \ =|J| dr \ d\theta[/tex] to find the area bounded by the curve [tex]x^{2/5}+y^{2/5}=a^{2/5}[/tex]

Attempt at a Solution:

I'm not really sure how to find the parametrization. Once I have that, calculating the Jacobian is simple. Then all that's left is computing a double integral. Right?

Hint: Your parametrization hints at polar co-ordinates.

##x = rcos\theta, y = rsin\theta##
 
  • #3
But if I use ##x=rcos\theta## and ##y=rsin\theta##, then I have ##dx=rd\theta## and ##dy=rd\theta##. Also ##|J|=r##. However we want ##dx \ dy=|J|dr \ d\theta##.

Using ##x=rcos\theta## and ##y=rsin\theta##, then the curve is given by [tex](rcos\theta)^{\frac{2}{5}}+(rsin\theta)^{\frac{2}{5}}=a^{\frac{2}{5}}[/tex]

So to find the area bounded by this curve we want to perform a double integral. I'm just confused on setting it up. Now that we're in terms of ##r## and ##\theta##, how do we figure out the limits of integration for the given curve?
 
Last edited:

1. How do you determine the appropriate parametrization for finding the area bounded by a curve?

The appropriate parametrization for finding the area bounded by a curve is determined by considering the properties of the given curve and the boundaries of the area. Generally, a parametrization that covers the entire curve and allows for easy integration is preferred. This can be achieved by choosing a suitable parameter, such as arc length or angle, and expressing the curve in terms of that parameter.

2. Can any parametrization be used to find the area bounded by a curve?

No, not all parametrizations are suitable for finding the area bounded by a curve. The chosen parametrization must cover the entire curve and allow for easy integration. Additionally, the parametrization should also have a one-to-one correspondence with the curve, meaning that each point on the curve should correspond to a unique value of the parameter.

3. How does the direction of parametrization affect the area bounded by a curve?

The direction of parametrization does not affect the area bounded by a curve. As long as the parametrization covers the entire curve and allows for easy integration, the direction does not make a difference in the final result. However, the direction of parametrization does affect the sign of the resulting area, so it is important to be consistent in the chosen direction.

4. What is the significance of the boundaries when finding the area bounded by a curve?

The boundaries are essential in finding the area bounded by a curve. They define the limits of integration and determine the area that is being calculated. It is important to properly identify and express the boundaries in terms of the chosen parameter in order to accurately calculate the area.

5. Are there any limitations to using parametrization to find the area bounded by a curve?

While parametrization is a commonly used method for finding the area bounded by a curve, it does have some limitations. For more complex curves or boundaries, it may be difficult to find a suitable parametrization. In these cases, other methods such as using Green's theorem or breaking the area into smaller, simpler pieces may be more effective.

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