The integral of interest is the total flux along one petal of the polar rose with 3 petals where the vector field F acting on the line described by the rose is the radial vector field.
The total flux is given by the flux integral `int_C vecF * vecn ds` and can be rewritten as `int_C vec F * d vecs` and further as `int_a^b vec F(vecs(t)) * vecs'(t) dt`
where `t` , `a <= t <=b` , is the parameter of the curve of interest.
The equation for the polar rose with 3 petals is `r(theta) = cos3theta` which can be rewritten as
`vecs(theta) = theta veci + cos3theta vecj`
Now the radial vector field `vecF(x,y) = << x, y >>` is described more simply in the polar coordinate plane by `vecF(theta,r) = 0 veci +r vecj ` . Due to this, and that the graph of interest is a polar rose , this problem is more easily dealt with in the polar coordinate system.
Now, `vecF (vecs(theta)) = 0 veci + cos3theta vecj`
Also, `vecs' (theta) = (d vecs(theta))/(d theta) = vec i - 3sin3theta vec j` , so that
`vec F(vecs(theta)) * vecs'(theta) = (0,cos3theta) * (1,-3sin3theta) `
`= -3cos3thetasin3theta` (the dot product of the two vectors).
If we want the flux integral over just one petalof the polar rose, then we can take `-pi/6 <=theta <= pi/6` which is the range of `theta` over the uppermost petal.
Then, the flux integral of interest `I` is given by
`I = int_a^b vecF(vecs(theta)) * vecs'(theta) d theta = -int_(-pi/6)^(pi/6) 3cos3thetasin3theta d theta`
Reparameterize and let `phi = 3theta` , `implies (d theta)/(d phi) = 1/3` then
`I = -int_(-pi/2)^(pi/2) cos phi sin phi d phi = cos^2 phi |_(-pi/2)^(pi/2) = 0 - 0 = 0`
Answer: the flux integral `I` = 0. The flux on one side of the petal negates the flux on the other as the petal starts and finishes at zero and is symmetric as regards the direction of the vector field.