0

Let $E$ be defined in $\mathbb{R}^3$ as the area between the parabola $z = x^2 + y^2$ and the plane $z = 2$. Given the vector $\vec{F} = \langle 2x, 2y, 0 \rangle$, find the flux integral $\iint_S \vec{F}\cdot\mathrm{d}\vec{r}$ using the divergence theorem.

So, we are looking for $$\iiint_E \text{div}\,\vec{F}\,\mathrm{d}V$$ Finding the divergence of $\vec{F}$ is easy, but I've been having trouble with the bounds. How do you set up the bounds for this integral?

Mailbox
  • 896
  • Do you really need to set up a triple integral, there is an easier way to calculate the required volume using a single integral. – WW1 Dec 11 '22 at 02:59
  • Do you have a picture of what this region looks like? It may help to sketch cross sections. – TomKern Dec 11 '22 at 03:59
  • @WW1 Yes, it needs to be a triple integral. I thought of this too—why not just think of it as the continuous sum of circles? But they are specifically asking for how to set up the integral. – Mailbox Dec 12 '22 at 14:44
  • @TomKern I do, and it left me more confused than when I started. – Mailbox Dec 12 '22 at 14:45
  • Can you work out what the shadow of the region is in the $xy$ plane? That should give you the $x,y$ bounds (or $r,\theta$ if you want cylindrical coordinates) – TomKern Dec 12 '22 at 18:10
  • You really should have learned how to evaluate a triple integral of a rather common shape before moving on to the divergence theorem. Anyway, $\iint_S F dS + \iint_D F \ dD = \iiint_E \nabla\cdot F \ dE,$ where $D$ is a disc in the plane $z = 2.$ That is the real meat of this exercise. To evaluate $\iiint_E \nabla\cdot F \ dE,$ I would use cylindrical coordinates. $\int_0^{2\pi}\int_0^{\sqrt 2}\int_{r^2}^2 r\ dz\ dr\ d\theta$ – user317176 Dec 13 '22 at 02:28
  • That last part was what I was missing! $x^2 + y^2 \leq z \leq 2 \implies r^2 \leq z \leq 2$. – Mailbox Dec 15 '22 at 21:03

1 Answers1

0

$$ V=\int_0^2 \int _{-\sqrt z} ^{\sqrt z} \int_{-\sqrt{z-y^2}}^{\sqrt{z-y^2}}dxdy dz $$

$$ V=2\int_0^2 \int _{-\sqrt z} ^{\sqrt z} \sqrt{z-y^2} dy dz $$ let $y =\sqrt z \sin \theta$ so that $$.$$ $dy = \sqrt z \cos\theta d \theta$ $$.$$and $\sqrt{z-y^2} = \sqrt z \cos \theta$

$$ V=2\int_0^2 z\int _ {-\frac \pi 2} ^{\frac \pi 2} \cos ^2 \theta d\theta dz $$ $$ \implies V = \pi \int_0^2 z\; dz = 2 \pi$$

Which is the same answer you get by integrating cross sections parallel to the $x-y$ plane, which are circles having radius $r = \sqrt z$ so each infinitesimal cylinder has volume $dV = \pi z dz$

WW1
  • 10,497