Finding the velocity of flow described by a vector field

Snoldermus
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Homework Statement


Consider the surface, S, in the xyz-space with the parametric representation: S: (, ) = [cos() , sin() , ] −1/2 ≤ ≤ 1/2 0 ≤ ≤ os().
The surface is placed in a fluid with the velocity potential f of a flow: = y*^2 + z*^2
a) Find the velocity of the flow described by a vector field (, , ).

b) What is the velocity at the point = (−1,5,0)

Homework Equations

The Attempt at a Solution



so this is the solution given for a and b, however i don't understand the answer at b. How can the different equations just be summed into one number, i would understand it if was squared--> then taken the sum of the numbers --> then taken the squareroot. Can anyone confirmed that this solution is wrong?[/B]
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Snoldermus said:
parametric representation: S: (, ) = [cos() , sin() , ] −1/2 ≤ ≤ 1/2 0 ≤ ≤ os().
Please clarify the surface S(,) and missing arguments for cosine, sine and the z coordinate.
 
Sry, i copied it and it messed it up a bit, thought i fixed it though here is a picture should be clear
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Velocity is a vector, so just saying that velocity is 16 seems to be wrong.

Also the magnitude of the velocity vector at that point is not 16.
 
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Likes scottdave
Thanks :) what i thought aswell
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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