# Evaluate Line integral

## 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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nrqed
Homework Helper
Gold Member

## 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?

Well isn't the right half of the circle just when x>0? So I thought it would correspond to all values of y...

nrqed
Homework Helper
Gold Member
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?

Ah, sorry. I mean to say -pi/2 to pi/2. I just fixed it.

nrqed