Line integral

[boneill3]

If I have a conservative vector F = e^x sin(y) + e^x cos(y)
over the path (0,0) to (-1,pi/2)

Is correct that I can just plu the values into the function to calculate the integral

ie

e^-1 sin(pi/2) + e^-1 cos(pi/2) - e^0 sin(0) + e^0 cos(0)


= e-1

regards

[jemidiah]

F is not a vector field, though you've called it a "conservative vector". You wouldn't use F to evaluate the integral anyway, you'd use its potential function; that is, the function whose gradient is F.

If you meant F = e^x sin(y) i + e^x cos(y) j for i and j being unit vectors, the potential function could be e^x sin(y) [derivation not shown], in which case the integral you seek would be

e^(-1) sin(pi/2) - e^0 sin 0= 1/e,

which follows from the Gradient Theorem.

[boneill3]

The question said show that the integral \int e^x sin(y) dx + e^x cos(y) dy is independent of the path and find its value?

I proved it was independent because the partial derivatives fpd/ypd = gpd/xpd

[jemidiah]

That's not what you typed in the first post, but it's about what I thought. I'm using < and > to indicate vectors, "dot" to mean dot product, and "grad" to mean gradient. You can say r=<x, y>, so dr = <dx, dy>, which will give the (IMO) more familiar form of the line integral from the article I linked, \int [vector field; gradient of phi in the article] dot dr. Doing this to your integral gives

\int e^x sin(y) dx + e^x cos(y) dy = \int <e^x sin(y), e^x cos(y)> dot <dx, dy> = \int <e^x sin(y), e^x cos(y)> dot dr.

You've already noted that the integrand, F = <e^x sin(y), e^x cos(y)>, is a conservative vector field, and I've already given the potential function, phi = e^x sin(y), which is also the phi from the article. Applying path independence, you get

\int <e^x sin(y), e^x cos(y)> dot dr = \int F dot dr = \int grad phi dot dr = phi(end) - phi(start) = phi(-1,pi/2) - phi(0,0) = e^(-1) sin(pi/2) - e^0 sin(0) = 1/e.


Just from your first post, it looks like you might want to make sure vector fields, scalar fields, and potential functions are clear and distinct in your own mind.

[boneill3]

Thankyou for your detailed explaination.
This is my first year at Multi-variable calculus So there are a lot of terms I haven't fully grasped yet.

thanks for the link it helped explain the potential function better than my lecture notes.
regards

[jemidiah]

No problem, glad it worked out.