Line integral

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

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.

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

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.

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

No problem, glad it worked out.