# flux of vector field through a paraboloid

I'm trying to find the flux of $$\mathbf{F} = x \, \mathbf{i} + (x^2+2z) \, \mathbf{k}$$ out of the space limited by the paraboloid $2z = 1 -x^2 -y^2$ and the $xy$-plane.

I've tried to parametrize from $r(x,y,z)$ to $r(x,y)$ and $r(x,z)$ but they yield impossible integrals to calculate. And at that point spherical/cylindrical coordinates doesn't help either.

Does any one have a good tip?

Cheers,

• a) What's xi? b) You can get your formulas typeset by enclosing them in single (inline) or double (displayed) dollar signs. – joriki Oct 14 '11 at 14:51
• Apply the divergence theorem. – Sasha Oct 14 '11 at 15:45

Let $\mathcal{R} = \{(x,y,z) \in \mathbb{R}^3 : 0 < 2 z < 1-x^2-y^2 \}$, and let $\partial \mathcal{R}$ denote its boundary would consist of parabolid $z = \frac{1}{2}(1-x^2-y^2)$ for $x^2+y^2<1$, and its lid $\{(x,y,z): z =0 \land x^2+y^2 < 1 \}$.
What you need to compute is $\int_{\partial \mathcal{R}} \mathbf{F} \cdot \mathrm{d} \mathbf{S}$. Using divergence theorem, and $F_x = x$, $F_y = 0$, and $F_z = x^2 + 2 z$:
$$\int_{\partial \mathcal{R}} \mathbf{F} \cdot \mathrm{d} \mathbf{S} = \int_\mathcal{R} (\mathbf{\nabla}\cdot \mathbf{F}) \mathrm{d} V = \int_\mathcal{R} \left( \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \right) \mathrm{d} V = \int_\mathcal{R} \left( 1 + 2\right) \mathrm{d} V = 3 V_\mathcal{R}$$
The volume, after switching to cylindrical coordinates $x = r \cos \phi$, $y = r \sin \phi$: $$V_\mathcal{R} = \int_0^{2 \pi} \mathrm{d}\phi \int_0^1 r \mathrm{d} r \int_0^{\frac{1-r^2}{2}} \mathrm{d} z = \frac{\pi}{4}$$