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Evaluate Line integral

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  • #1
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Homework Statement


Evaluate the line integral, where C is the given curve.
integral of xy^4. C is the right half of the circle x^2 + y^2 =16

Homework Equations


integral of line integral= integral of r(t) |r'(t)| dt


The Attempt at a Solution


I set up a parametric equation to be r(t)=(4cost, 4sint)
then r'(t)=(-4sint, 4cost) --> |r'(t)|=4
My t is between -pi/2 and pi/2---these are my limits of integration.

With the limits of integration, my integrand is (4)(cost)(sint)^4. Setting my u=sint, my final answer came out to be 8/5. I know I got the wrong answer though...
 
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Answers and Replies

  • #2
nrqed
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Homework Statement


Evaluate the line integral, where C is the given curve.
integral of xy^4. C is the right half of the circle x^2 + y^2 =4

Homework Equations


integral of line integral= integral of r(t) |r'(t)| dt


The Attempt at a Solution


I set up a parametric equation to be r(t)=(4cost, 4sint)
then r'(t)=(-4sint, 4cost) --> |r'(t)|=4
My t is between -pi/4 and pi/4---these are my limits of integration.

With the limits of integration, my integrand is (4)(cost)(sint)^4. Setting my u=sint, my final answer came out to be 8/5. I know I got the wrong answer though...
Why did you use pi/4 and -pi/4?? If you you plug those values in your parametrized curve, you don't get the limits corresponding to the right half of a circle! Or am I missing something?
 
  • #3
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Well isn't the right half of the circle just when x>0? So I thought it would correspond to all values of y...
 
  • #4
nrqed
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Well isn't the right half of the circle just when x>0?
Right
So I thought it would correspond to all values of y...
Your parametrized curve is 4 cos(t), 4 sin(t)

If you plug t = pi/4 in that what do you get?
 
  • #5
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Ah, sorry. I mean to say -pi/2 to pi/2. I just fixed it.
 
  • #6
nrqed
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Ah, sorry. I mean to say -pi/2 to pi/2. I just fixed it.
Ok. The other error is that the radius is 2, not 4. So you must changed your parametrized curve
 

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