Surface Integrals: Finding Flux

Question

Find the flux across s. If S is closed use the positive(outward) orientation.

$F(x,y,z) = \langle 9x, 9y,4\rangle$

$S$ is part of the paraboloid $z = x^2 + y^2$ between planes $z=0$ and $z=4$; $n$ points downwards.

Any help would be appreciated. :)

Answer 1

So you want to find Flux = $\iint (F \cdot n) d\sigma$. Note that the (non-unit) normal at any point is $<2x, 2y, -1>$ (from the gradient) which is definitely the downward pointing normal since $z \leq 0$. So $n = \frac{<2x,2y,-1>}{||<2x,2y,-1>||}$. Then $F \cdot n = \frac{18x^2 + 18y^2 -4}{||<2x,2y,-1>||}$. Recall that $ d\sigma = ||<2x,2y,-1>|| dA$, so the integral turns into $\iint_R (18x^2 + 18y^2 -4) dA$. We just need to determine the region of integration. The projection of the paraboloid onto the $xy$ plane is the area enclosed by the circle $x^2 + y^2 = 4$. For the sake of calculation, let us describe this in polar coordinates. Then this is $$\int_0^{2\pi} \int_0^2 (18r^2 -4)r dr d\theta$$ which if you evaluate is $128\pi$.

Answer 2

We can parametrize the surface $S$ as follows: $$ \mathbf{X}(s,t) = \langle s, t, s^2 + t^2 \rangle \qquad\qquad \text{for } s^2+t^2 \leq 4$$ This parametrization takes the region: $$ D = \{(s,t) \in \mathbb{R}^2 \mid s^2 + t^2 \leq 4\} $$a solid disk of radius 2 in $s$-$t$ space, and maps its to the surface $S$ in $\mathbb{R}^3$

We see that: \begin{align*} \frac{\partial \mathbf{X}}{\partial s} &= \langle 1, 0, 2s \rangle \\ \frac{\partial \mathbf{X}}{\partial t} &= \langle 0, 1, 2t \rangle \end{align*} So the normal vector for our surface integral (adding a $\pm$, so we choose the orientation pointing downward) is: \begin{align*} \mathbf{N}(s,t) &= \pm \frac{\partial \mathbf{X}}{\partial s} \times \frac{\partial \mathbf{X}}{\partial t} \\ &= \pm\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 2s \\ 0 & 1 & 2t \end{vmatrix} \\ &= \langle 2s, 2t, -1 \rangle \end{align*}

Given the vector field: $$ \mathbf{F}(x,y,z) = \langle 9x, 9y, 4 \rangle $$ We get the flux through the surface $S$ is found expressed by the surface integral: \begin{align*} \iint_S \mathbf{F} \cdot d\mathbf{S} &= \iint_D \mathbf{F}(\mathbf{X}(s,t)) \cdot \mathbf{N}(s,t) \, dA \\ &= \iint_D \langle 9s, 9t, 4 \rangle \cdot \langle2s,2t,-1 \rangle \, dA \\ &= \iint_D (18s^2+18t^2-4) \, dA \end{align*} Since the region $D$ is a solid disk of radius 2, we change from Cartesian coordinates $s,t$ to polar coordinates $r,\theta$, noting that: $$ dA = ds \, dt = r \, dr \, d\theta $$ so the surface integral simplifies to: \begin{align*} \iint_S \mathbf{F} \cdot d\mathbf{S} &= \iint_D (18s^2+18t^2-4) \, dA \\ &= \int_0^{2\pi} \int_0^2 (18r^2 - 4) r \, dr \, d\theta \\ &= 2\pi \int_0^2 (18r^3 - 4r) \, dr \\ &= 2\pi \left[\frac{18}{4}r^4 - 2r^2 \right]_0^2 \\ &= 2\pi \left[(18)(4) - 8 \right] \\ &= 128\pi \end{align*}