Hi all. I'm having troubles with a problem from my Calculus IV course. The problem states:
I began by parameterizing the line I was given. For any given y value, x will be three times that value, so I said that r(t)=<3t,t>. From there I took the derivative and found its magnitude.
r′(t)=<3,1> and ∣∣r′(t)∣∣=32+12=10
I then converted the given function into a function of t. x(t) = 3t and y(t) = t, so e^(xy) = e^(3t^2). That means the integral I'm really solving is:
∫−11e3t2⋅10dt=10⋅∫−11e3t2dt
But here's where I get stuck, because I don't know how to integrate that function. Wolfram Alpha says the answer involves the "Imaginary Error Function," but I don't know what that is, much less what its value might be. I'm thinking I messed up somewhere, because I don't think the textbook would ask me to integrate something like that. However, I can't see any step where I might have made an error. Any help would be much appreciated.
For exercises 21-28, evaluate the multivariate line integral of the given function over the specified curve.
22) f(x,y)=exy with C the line x = 3y for y∈[−1,1]
I began by parameterizing the line I was given. For any given y value, x will be three times that value, so I said that r(t)=<3t,t>. From there I took the derivative and found its magnitude.
r′(t)=<3,1> and ∣∣r′(t)∣∣=32+12=10
I then converted the given function into a function of t. x(t) = 3t and y(t) = t, so e^(xy) = e^(3t^2). That means the integral I'm really solving is:
∫−11e3t2⋅10dt=10⋅∫−11e3t2dt
But here's where I get stuck, because I don't know how to integrate that function. Wolfram Alpha says the answer involves the "Imaginary Error Function," but I don't know what that is, much less what its value might be. I'm thinking I messed up somewhere, because I don't think the textbook would ask me to integrate something like that. However, I can't see any step where I might have made an error. Any help would be much appreciated.