Line integral over plane curve

  • Thread starter DWill
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Homework Statement


Integrate f(x,y) = (x^3)/y over the curve C: y = (x^2)/2, 0 <= x <= 2


Homework Equations





The Attempt at a Solution


So far I'm only familiar with line integrals over space curves such as questions like this: Find the line integral of f(x,y,z) = x + y + z over the straight-line segment from (1, 2, 3) to (0, -1 1). Here I know to find equations for x(t), y(t) and z(t) based on the points given. This gives me r(t), which I take derivative to find |v(t)|. To find f in terms of t I substitute x(t), y(t), z(t) into the equation for x, y, and z, then finally solve the integral of f(t) * |v(t)| dt over the interval.

I'm not sure how to start this question though when I'm given a plane curve? Thanks for any suggestions
 

Answers and Replies

  • #2
Defennder
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Your first step here, as in your stated case of evaluating line integrals for straight line paths would be to first parametrize the curve in terms of t. Take for example, the line y=2x + 3.

A parametric form of that graph would r(t) = ti + (2t+3)j. In this case, x is treated as a parameter t which then varies over the course of the line in space. When that is done the line integral of f(x,y) over the path defined by 2x+3 is given by [tex]\int^{t_b}_{t_a} f(t) \left| \frac{dr}{dt} \right| dt[/tex], where t_a and t_b are the respective values of t corresponding to the endpoints of the graph over which the line integral is performed. Express f(x,y) in terms of r(t) to get f(t) and the rest follows.
 
  • #3
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Thanks, I tried doing as you said: I made r(t) = ti + (t^2/2)j, and then found |v(t)| = sqrt(1 + 4t^2). Then I got f(t) = 2t. Finally I took the integral of 2t * sqrt(1 + 4t^2) dt over 0 <= t <= 2, but don't end up with the correct answer. Where did I mess up here? To express f(x,y) in terms of r(t) as you said I just plug in x(t) and y(t) from the r(t) equation for x and y, right?

thanks again
 
  • #4
Defennder
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Thanks, I tried doing as you said: I made r(t) = ti + (t^2/2)j, and then found |v(t)| = sqrt(1 + 4t^2).
It's not 4t^2 here.
 
  • #5
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Ohh I see..stupid mistake
 
  • #6
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Homework Statement


Ok this is a related question, didn't want to make a new topic:

A circular wire hoop of constant density d lies along the circle x^2 + y^2 = a^2 in the xy-plane. Find the hoop's moment of inertia about the z-axis.

Homework Equations


I_z = Integral of (x^2 + y^2) * d ds, over a curve C

The Attempt at a Solution


How do I parametrize this equation? I thought of something like r(t) = (sin t)i + (cos t)j, but that doesn't take into account radius a?
 
  • #7
Defennder
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You're partially right here. What then should you do to make r(t) = sin (t) i + cos (t) j take the radius of the circle into account? Remember the coordinate transformation equations for x,y in terms of r,theta.
 

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