**Part a)**

The integral of interest is the ** total flux** along

**where the vector field F acting on the line described by the rose is the**

*one petal of the polar rose with 3 petals*

**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

**, this problem is more easily dealt with in the polar coordinate system.**

*polar rose*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 petal**of 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.**

The radial vector field is given by the equation `bar F = x*bar i + y*bar j.`

You need to evaluate the flux integral `int_C F*n*ds` for the radial vector field `bar F = x*bar i + y*bar j` and the polar curve `r = cos 3theta.`

Since the problem provides the information that you need to evaluate the flux integral of only one petal of polar curve r = cos 3 theta, hence, the loop created has the initial and end points equal, thus, the flux integral is equal to 0, such that:

`int_C F*n*ds = int_C bar F*(grad f)/||grad f||*||gradf||*ds = 0.`

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.