0

I have to evaluate

$$\iint_{S}\vec{F}\cdot d\vec{S}$$

with $$\vec{F}(x,y,z)=(y,-x,z)$$ and $S$ is bounded by $z=9-x^2-y^2$ and $z=5$. I can't use Gauss Theorem to do it.

I thought about using the following "formula" $$\int_{S}\vec{F}\cdot d\vec{S}=\iint_{D}\left(-P\cdot\frac{\partial g}{\partial x}-Q\cdot\frac{\partial g}{\partial y}+R\right)dA$$ with $\vec{F}=(P,Q,R)$ and $S$ is the graph of $z=g(x,y)$.

However, as $z\geq5$, do I have to take $$g(x,y)=9-x^2-y^2-5?$$

mvfs314
  • 2,017
  • 15
  • 19
  • 1
    IDK about the formula you mention, but you can parameterize both pieces of $S$ in cylindrical coordinates easily enough. See this wiki article for guidance if you're unfamiliar with this process. – user170231 Dec 20 '23 at 16:26
  • @user170231 I tried to parameterize as follows: $r(u,v)=(u\cos v,u\sin v,9-u^2)$. But when I evaluate it, the final answer is $28\pi$. If I use Gauss theorem, I get $8\pi$. – mvfs314 Dec 20 '23 at 16:29
  • 1
    That's right, what you found is the integral along the paraboloid. You also need to consider the integral along the disk (the other "piece" of $S$ I alluded to earlier). – user170231 Dec 20 '23 at 16:30
  • Yes, what I did was: $0\leq u\leq 4$ and $0\leq v\leq2\pi$. But why the answers are different? – mvfs314 Dec 20 '23 at 16:32
  • 1
    (1) $9-x^2-y^2=5 \implies x^2+y^2=2^2$ so you should have $0\le u\le2$. (2) I wouldn't call your answer ($28\pi$) "different", rather it's incomplete. You want to integrate $F$ across the entire surface, not just the paraboloidal part. One would find the disk contributes $-20\pi$ so that the overall integral is indeed $8\pi$. – user170231 Dec 20 '23 at 16:37
  • I always forget these disks. The minus sign is because the normal vector on the disk is downward? – mvfs314 Dec 20 '23 at 16:48
  • 1
    It's not explicitly mentioned but based on the expected answer, it's a consequence of the orientation chosen for the surface. You want the normal to $S$ to point outward everywhere on $S$. $S$ lies above the plane $z=5$, so relative to that plane, outward means downward. – user170231 Dec 20 '23 at 16:52

1 Answers1

1

Parameterize the paraboloid with

$$\vec s_P = (u \cos v, u \sin v, 9-u^2), \quad u\in[0,2] \land v \in[0,2\pi]$$

and normal vector

$$\vec n_P = \frac{\partial \vec s_P}{\partial u} \times \frac{\partial \vec s_P}{\partial v} = \left(2u^2\cos v,2u^2\sin v,u\right)$$

and the disk with

$$\vec s_D = (u \cos v, u \sin v, 5), \quad u\in[0,2] \land v\in[0,2\pi]$$

and normal (easy to determine by inspection) $\vec n_D = (0,0,-1)$.

Integrating along the respective surfaces yields

$$\begin{align*} \iint\limits_{\rm paraboloid} \vec F \cdot d\vec S &= \int_0^{2\pi} \int_0^2 \left(9u-u^3\right) \, du \, dv \\[2ex] \iint\limits_{\rm disk} \vec F \cdot d\vec S &= \int_0^{2\pi} \int_0^2 -5u \, du\,dv \\[2ex] \implies \iint_S \vec F\cdot d\vec S &= \int_0^{2\pi} \int_0^2 \left(4u-u^3\right) \, du \, dv \end{align*}$$

which agrees with your solution via divergence theorem.

user170231
  • 19,334