Author Topic: Vector Calculus (Read 8188 times) Tweet Share

QuantumJG · « **Reply #30 on:** November 04, 2010, 11:25:19 am »

Hey guys,

I keep getting this question wrong.

Let a surface be paramaterised by

Φ(u,v) = (u-v,u+v,uv)

Let D be a unit disc (I'm assuming u² + v² $\le$ 1)

Find the area of Φ(D).

I keep getting $\dfrac{2 \sqrt{2} \pi}{3}$ whereas the answer is $\dfrac{ \pi }{3} (6 \sqrt{6} - 8)$

/0 · « **Reply #31 on:** November 04, 2010, 12:05:03 pm »

$\mathbf{r}_u \times \mathbf{r}_v=(u-v)\mathbf{i}-(u+v)\mathbf{j}+2\mathbf{k}$

What we want is $\int_{disk} |\mathbf{r}_u \times \mathbf{r}_v| \,dA=\int_{disk} \sqrt{2u^2+2v^2+4}\,dA$

Now switching to polar works well:

$\int_0^{2\pi}\int_0^1 s\sqrt{2s^2+4}\,ds\,d\phi$

QuantumJG · « **Reply #32 on:** November 04, 2010, 03:19:56 pm »

I realized now where I stuffed up.

I did basically what you did, but after that to simplify things I let α = s² + 2 but didn't change my terminals!

QuantumJG · « **Reply #33 on:** November 06, 2010, 06:39:55 pm »

Another question:

F = (x-y,y-z,z-x)

Evaluate the surface integral of F if the surface S is the closed region:

x² + y² $\le$ z $\le$ 1 & x $\ge$ 0

Using the divergence theorem my answer was $\dfrac{3 \pi}{2}$ whereas the answer is $\dfrac{3 \pi}{4}$

/0 · « **Reply #34 on:** November 06, 2010, 08:14:04 pm »

In cylindrical coordinates, $s^2=x^2+y^2$ ,

$\int_0^1\int_{s^2}^1\int_0^\pi 3s\,d\phi\,dz\,ds$

QuantumJG · « **Reply #35 on:** November 06, 2010, 09:22:58 pm »

Ok now I get it. I stuffed up my terminals for z. Thanks /0.

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