Use the divergence theorem to compute flux integral $ \iint SF⋅dS$, where $F(x,y,z)=yj−zk $ and $S$ consists of the union of paraboloid $ y=x^2+z^2,0≤y≤1$, and disk $ x^2+z^2≤1,y=1$, oriented outward. What is the flux through just the paraboloid?
Here's what I have so far: enter image description here
The answer key says it's $-\pi$ whereas I am getting $\pi$.
If $\vec{F}(x,y,z)=\langle0,y,-z\rangle$ then $\nabla _{\{x,y,z\}}\cdot \vec{F}(x,y,z)=0$
Hence, the total flux is 0. We must use the surface integral.
$$\vec{r}=\langle x,x^2+z^2,z \rangle$$
$$\vec{\delta_x}= \delta_x \cdot \vec{r} = \langle 1, 2x, 0 \rangle$$ $$\vec{\delta_z}= \delta_z \cdot \vec{r} = \langle 0, 2z, 1 \rangle$$
The normal vector for the paraboloid: $$\vec{N}=\vec{\delta_x}\times \vec{\delta_z}=\langle 2x,-1,2z\rangle$$
Integrand for the surface integral: $$\vec{F}(x,y,z)\cdot\vec{N}=-y-2 z^2|_{y=x^2+z^2}=-x^2 - 3 z^2$$
$$\int _{-1}^1\int _{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\left(-x^2-3 z^2\right)dzdx=\int_{-1}^1 \left(-2 \left(1-x^2\right)^{3/2}-2 x^2 \sqrt{1-x^2}\right) \, dx=-\pi$$