v$(r,φ,θ) = (r cos2 θ)$r$ − (rcosθsinθ)$θ$ + 3r$φ, where r, θ and φ are the unit spherical vectors.
I was trying to calculate the line integral of the function along the path described in the picture below, but I need the function in cartesian coordinates to do so right? If I do, how do we do that? 
It's simpler to work this out in spherical coordinates. I assume that $\theta$ is the polar angle. You can verify that $$\nabla \times \mathbf v = (3 \cot \theta) \boldsymbol r - 6 \boldsymbol \theta + (\sin 2 \theta) \boldsymbol \phi, \\ (0, 0, 1) \big\rvert_{\theta = \pi/2} = -\boldsymbol \theta, \\ (1, 0, 0) \big\rvert_{\phi = \pi/2} = -\boldsymbol \phi,$$ therefore you need to evaluate $\iint 6 \, dS$ over the shaded quarter-disk and $-\iint \sin 2 \theta \, dS$ over the shaded triangle, the latter giving $$-\int_{\arctan(1/2)}^{\pi/2} \int_0^{\csc \theta} r \sin 2 \theta \, dr d\theta = -\int_{\arctan(1/2)}^{\pi/2} \cot \theta \, d\theta = -\ln \sin \theta \bigg\rvert_{\theta = \arctan(1/2)}^{\pi/2}.$$