Thread: [SOLVED] Calc III Line Integrals

Hello, I can't figure this one out. Can someone please help me with this problem?


Evaluate the integral



along the stated curve

a.) y = sin((pi*x)/2) from (0,0) to (1,1)

b.) x = y^3 from (0,0) to (1,1)

See attachment

Thanks for the quick response.

I did that for a. but the answer is supposed to be 3.

Note the fied is conservative with

phi = 3/2*x^2 +2xy -y^2/2

so we have path independence

The answer to both a and b are 3

There must be a mistake in your integration as even working out the integral in my first post yields 3 as well.

Oh man... I don't get how you got 3.

Originally Posted by Calculus26

Note the fied is conservative with

phi = 3/2*x^2 +2xy -y^2/2

so we have path independence

The answer to both a and b are 3

How did you get phi? and that we have path independence?

See attachment for details of integration.

To answer your second question have you studied Path Independence Yet?

Let the field be F = f i + g j

if df/dy = dg/dx F is the gradient of a scalar function phi.

The line integral is then simply phi(endpoint) - phi(initialpoint)

If you haven't studied this you will including how to find phi -- note for the phi I gave grad(phi) = F

and phi(1,1)- phi(0,0) = 3

For a complete discussion of path independence and conservative fields

see my web site Line Integrals

No, we haven't yet. But I'm taking Calc III in summer school and everything moves so fast. I'm sure I'll be learning it tomorrow. Thank you sooo much. You were very helpful!