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Problem - need help for part (ii)

Let $\vec{F} = y \vec{i} -x \vec{j} + z \vec{k}$ and let the surface $S$ be the part of the paraboloid $ z = 4 - x^2 - y^2$ with $z \geq 0 $, oriented with $\vec{n}$ upwards. Calculate the flux integral $\int_S \vec{F} \cdot d\vec{S}$ using

i) Cartesian coordinates

ii) cylindrical coordinates


My attempt for (i)

We parameterize the surface $S$ using Cartesian coordinates, $$\vec{r}(u,v) = (u,v, 4 - u^2 - v^2), $$ and let $f(x,y,z) = x^2 + y^2 + z$. The normal to the surface $S$, $\vec{n}$, is found by computing grad(f), i.e. $$\vec{n} = \nabla f(x,y,z) = (2x,2y,1) = (2u,2v,1). $$ Time to evaluate the integral, \begin{align*} \int_S \vec{F} \cdot d\vec{S} & = \int_S \vec{F}(\vec{r}(u,v)) \cdot \vec{n} dS \\ & =\int_S (v,-u,4-u^2-v^2) \cdot (2u,2v,1) dS \\ & = \int_S (4 - u^2 - v^2) dS. \tag{1} \end{align*} Since the surface is defined as a paraboloid, we will use cylindrical coordinates to find out the limits of integration, \begin{align} u &= r\cos(\theta), \\ v &= r\sin(\theta), \\ z &= z. \end{align} As $z \geq 0$, then we have $u^2 +v^2 \leq 4$ which implies that the $ 0 \leq r \leq 2$.

Now the integral becomes, $$\int_0^{2 \pi} \int_0^2 (4-r^2)r dr d\theta = \dots = 8\pi. $$

A predicament - SOLVED - See first comment below.

Let's come back to $(1)$. To find the limits, I originally did this:

As $z \geq 0$, we have $u^2 + v^2 \leq 4$ which implies that $$-\sqrt{4-u^2} \leq v \leq \sqrt{4-u^2} \\ 0 \leq u \leq 2.$$ Putting this into the integral gives an answer of $4\pi$. What was wrong with this reasoning for finding the limits of integration?


For part (ii)

I'm going to skip the detail but using polar coordinates gives, $$ \vec{n} = (2rcos(\theta), 2rsin(\theta),r) \\ \vec{F}(\vec{x}(r,\theta)) = (rsin(\theta), -rcos(\theta), 4-r^2). $$

Substitute the findings into the integral, \begin{align*} \int_S \vec{F} \cdot \vec{n} dS & = \int_S (4-r^2) dS \end{align*}

Here is the problem, what is $dS$? If I understand correctly, it was $dxdy$ but after the cylindrical coordinates parameterization, it changes to $drd\theta$. Then does that mean I'm missing the Jacobian,$\left| \dfrac{\partial (x,y)}{\partial (r,\theta)}\right| = r$ since I changed the variables?

If so, if gives the answers. I feel that I have reasoned correctly. Could anyone please point out any errors in my reasoning or confirm if I'm correct?

Many thanks

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  • $\begingroup$ For your predicament, notice that your bound on $u$ is incorrect; there's no reason why $u$ cannot take negative values. The correct bound is $|u|\leq 2$ i.e. $-2\leq u \leq 2$. $\endgroup$ – FH93 Jun 7 '15 at 11:03
  • $\begingroup$ @FH93 Wow, what a miss! Thanks for solving a logical error. $\endgroup$ – Jester Tran Jun 7 '15 at 11:05
  • $\begingroup$ math.stackexchange.com/questions/1312241/… $\endgroup$ – Mann Jun 7 '15 at 11:53
  • $\begingroup$ @Mann Thanks, that really helps. $\endgroup$ – Jester Tran Jun 7 '15 at 12:06
  • $\begingroup$ No problem, ^^. $\endgroup$ – Mann Jun 7 '15 at 12:07
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I'm not sure why you switched from x,y to u,v but it doesn't really matter. The surface is "the part of the paraboloid z=4−x^2 −y^2 with z≥0" so its boundary is the circle x^2+ y^2= 4 in the xy-plane. In Cartesian Coordinates, you can take x going from -2 to 2 and, for each x, y from -sqrt{4- x^2} to sqrt{4- x^2}. In polar coordinates r goes from 0 to 2 and $\theta$ from 0 to $2\pi$.

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  • $\begingroup$ Thanks, but I've already established the bounds of integration. I switched to u and v because I'm used to u and v being my parameters. My question now is part ii. $\endgroup$ – Jester Tran Jun 8 '15 at 1:41

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