How Do You Convert f=xy Into a Function of t for Integration Over a Curve?

[dylanhouse]

Homework Statement

I am asked to find the integral over C of fds given C={r=cos(2t), theta=2t, for 0<=t<=pi/2} and f=xy.

Homework Equations

The Attempt at a Solution

I know the integral over C of fds is the integral over C of [fsqrt(r'^2+r^2(theta)'^2)]dt, but I don't know how to convert my function f=xy into a function of t so that I can integrate using the dt. I know x=rcos(theta) and y=rsin(theta), but that doesn't allow me to integrate with respect to t.

 

[TheFerruccio]

You're basically asked to do a weighted line integral. It is similar to finding the length of C (the "line" in question), except, instead of f=1, you have f=xy.

So, you want to convert x and y to r and θ. Then, r and θ will be expressed in terms of t.

Do you know how to convert x and y to polar coordinates?

If you get that far, then you will be able to convert the polar coordinates using the "C={..." statement you gave.

 

[dylanhouse]

Perfect! I had converted x and y to polar coordinates but I was getting stuck with theta's and r's! I didn't realize that from how C was defined I could rewrite the r and theta's! Thanks a bunch.

 

[TheFerruccio]

You're welcome! That was usually my biggest hang-up with vector calculus was realizing the interplay between the functions and their domain, and how the domain of the function could be realized through simple direct substitution (say, x=f(t) and y=g(t) then integrate over t)