Question statement: Let $W$ be the 3D region under the graph of $f(x,y) = \exp(x^2+y^2)$ over the region in the plane defined by $1 \leq x^2 + y^2 \leq 2$. Find the flux of $\mathbf{F} = (2x-xy,-y,yz)$ out of the region W and check your answer using the divergence theorem.
My attempt: The divergence theorem is the easier method. Using it, we get $$\iint_{\partial W} \mathbf{F} \cdot d\mathbf{S}=\iiint_W \nabla \cdot \mathbf{F} dV = \iiint_W 1 dV = \int_1^{\sqrt 2} \int_0^{2\pi} re^{r^2} d\theta dr = \pi(e^2-e).$$
The direct computation is more difficult and I think that there are 3 surfaces which bound the volume $W$.
The first is the surface defined by $z=e^{x^2+y^2}$ over the region in the plane. I parametrise it using $\mathbf{r}(x,y) = (x,y,\exp(x^2+y^2)).$ I find it is easier to work with cylindrical polars, so I instead use $\mathbf{r}(r, \theta)=(r\cos\theta, r\sin \theta, e^{r^2}), 1 \leq r \leq \sqrt{2}, 0 \leq \theta \leq 2 \pi.$ I compute the normal $$\mathbf{N} = \mathbf{r}_r \times \mathbf{r}_{\theta} = (\cos \theta, \sin \theta, 2re^{r^2}) \times (-r\sin \theta, r \cos \theta, 0) = r(-2r \cos \theta e^{r^2}, -2r e^{r^2} \sin \theta, 1).$$
I check and this normal seems to be pointing outside the surface. Proceeding, we get $$\mathbf{F} \cdot \mathbf{N} = -r^2 e^{r^2} \sin \theta - 2e^{r^2} r^2 (2r \cos \theta - r^2 \sin \theta \cos \theta) + 2e^{r^2} r^3 \sin ^2 \theta.$$
Then $$\int_1^{\sqrt 2} \int_0^{2\pi} \mathbf{F} \cdot \mathbf{N} d\theta dr = -\pi e^2.$$
The second surface is the disk on the plane, i.e $x^2 + y^2 = r^2$ for $0 \leq r \leq \sqrt{2}.$ But there, $z = 0$ and the normal points in the $z$-direction and so the integral is $0$.
The third surface is the cylinder. I parametrise using $$\mathbf{r}(\theta, z) = (\sqrt{2} \cos \theta, \sqrt{2} \sin \theta, z), 0 \leq \theta \leq 2 \pi, 0 \leq z \leq e^2.$$ Going through the steps again we get $\mathbf{N} = \mathbf{r}_{\theta} \times \mathbf{r}_z = (\sqrt{2} \cos \theta, \sqrt{2} \sin \theta, 0).$ Then we obtain that $$\mathbf{F} \cdot \mathbf{N} = 4 \cos^2 \theta - 2\sqrt{2} \cos \theta \sin \theta - 2 \sin^2 \theta.$$
So we can get that $$\int_0^{2\pi} \int_0^{e^2} \mathbf{F} \cdot \mathbf{N} dz d\theta = 2 \pi e^2.$$
But then the sum is just $\pi e^2$ instead of $\pi e^2 - \pi e$.
I understand that the Divergence Theorem gives the answer quite easily, but I would be very grateful if someone could tell me where I went wrong here. Thank you very much for helping me.
There is a fourth surface: an internal cylinder with radius $1$: $$\mathbf{r}(\theta, z) = (\cos \theta,\sin \theta, z), 0 \leq \theta \leq 2 \pi, 0 \leq z \leq e.$$ Then $\mathbf{N} = \mathbf{r}_{\theta} \times \mathbf{r}_z = (\cos \theta, \sin \theta, 0)$ and $\mathbf{F} \cdot \mathbf{N} = 2 \cos^2 \theta - \cos \theta \sin \theta - \sin^2 \theta$. Hence the flux accross this fourth surface with the correct orientation, i.e. pointing towards the $z$-axis, is $$-\int_0^{2\pi} \int_0^{e} \mathbf{F} \cdot \mathbf{N}\, dz d\theta = -\pi e.$$ Finally, the total sum is $$\iint_{\partial W} \mathbf{F} \cdot d\mathbf{S}=-\pi e^2+0 +2\pi e^2-\pi e=\pi(e^2-e).$$ as we expected.