I'm working on a problem and solved both line integral and double integral to find circulation for verifying Stokes' Theorem.
I have a problem though, my answers for both methods DO NOT MATCH, so I obviously went wrong somewhere, but I can't locate my mistake (or my understanding might be off). Help is appreciated in pointing me back towards the right direction!
S is the part of the paraboloid $z = 9 - x^2 - y^2$ that lies above the plane $z = 5$ with upward orientation and $\vec{F} = \langle yz,-xz,xy \rangle$.
LINE INTEGRAL METHOD:
$\vec{r}(t) = \langle 2cost, 2sint, 5 \rangle$
$\vec{r}\text{ }'(t) = \langle -2sint, 2cost, 0 \rangle$
$\vec{F}(\vec{r}(t)) = \langle 10sint, -10cost, 4sintcost \rangle$
$\vec{F}(\vec{r}(t)) \cdot \vec{r}\text{ }'(t) = \langle 10sint, -10cost, 4sintcost \rangle \cdot \langle -2sint, 2cost, 0 \rangle = -20$
$\text{Circulation on S } = \int_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi} -20dt = -40\pi$
DOUBLE INTEGRAL METHOD:
$\gamma = \langle rcos\theta, rsin\theta, 4-r^2 \rangle$
$\gamma_r = \langle cos\theta, sin\theta, -2r \rangle$
$\gamma_\theta = \langle -rsin\theta, rcos\theta, 0 \rangle$
$\gamma_r \times \gamma_\theta = \langle 2r^2cos\theta, 2r^2sin\theta, r \rangle$
$\text{curl }\vec{F} = \langle 2rcos\theta, 0, 2r^2 - 8 \rangle$
$\int\int_R \text{curl }\vec{F} \cdot (\gamma_r \times \gamma_\theta)dA = \int_0^{2\pi}\int_0^2\langle 2rcos\theta, 0, 2r^2 - 8 \rangle \cdot \langle 2r^2cos\theta, 2r^2sin\theta, r \rangle drd\theta = 0$
Thus, the two methods do not equal the same, and I'm not sure where I went wrong.