Line integral where a vector field is given in cylindrical coordinates

[goohu]

Homework Statement

see image

Relevant Equations

equation of circles; radius = x^2 + y^2

What I've done so far:

From the problem we know that the curve c is a half-circle with radius 1 with its center at (x,y) = (0, 1).

We can rewrite x = r cos t and y = 1 + r sin t, where r = 1 and 0<t<pi. z stays the same, so z=z.

We can then write l(t) = [x(t), y(t), z ] and solve for dl/dt.

The integral can be rewritten as integral A dot l(t) dt, with the limits as 0 and pi.Now everything would be fine if the vector field A was given in cartesian coordinates but its not. You could transform different coordinate systems but I can't figure it out. Could someone please show me how to start on the last steps?

I know how to transform specific coordinates but I'm having trouble transforming a whole function. If we can express A in cartesian form then we can use scalar multiplication in the last step to solve the problem.

 

[haruspex]

We can rewrite x = r cos t and y = r sin t, where r = 1 and 0<t<pi.

No, that does not put the centre of the arc at (0,1,0). Did you mean y=1+ r sin(t)?

 

[goohu]

yes, my bad.

 

[goohu]

I made a mistake during the parameterization.
t should be : $-\pi / 2 < t < \pi / 2$

So the limits should be $\int_\frac{-\pi}{2}^\frac{\pi}{2} A(t) ⋅ l(t) \ dt$

 

[rude man]

EDIT: scratch my last post. Maybe should stay in cylindrical. Still don't see why parametrize.

 

[rude man]

I have my doubts about this one but I got zero for an answer.
EDIT: I think I at least have the right procedure even if I'm capable of math errors: $$ \int \mathbf A \cdot d\mathbf l = \frac { (8z + 2)} {3} $$.