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Integration of the Stream Function & Velocity Potential

  1. Dec 14, 2015 #1

    How did they go from ∂Ψ/dθ = m to Ψ = mθ + f1(r) ?

    I mean I understand how they got mθ but can someone explain the f1(r) term?

    Similarly, how did they go from -∂Ψ/dr = 0 to Ψ = c1 + g1(θ) ?

    I understand how they got c1 but why the additional term?

    I can't provide further working out as I think this is supposed to be pretty simple, I'm just missing something..

  2. jcsd
  3. Dec 14, 2015 #2


    Staff: Mentor

    Moved this thread out of the HW sections. Note that if you post in the HW sections, you must include the homework template.
    Ψ is assumed to be a function of both θ and r, so when you integrate ∂Ψ/dθ with respect to θ, you have to include a "constant" of integration that is a function of r alone. As a check, take the partial derivative with respect to θ of the right side, and you should get what they have on the left side.
    See above, with the difference that now the integration is with respect to r.
  4. Dec 15, 2015 #3
    That makes perfect sense. One question though, in this case we know that Ψ is a function of both θ and r but what if we were not aware of this? So say we were told to integrate ∂Ψ/dθ = m without being informed that Ψ is a function of both θ and r?

    Also for:


    How do we know that Ψ is a function of both x and y?

  5. Dec 15, 2015 #4


    Staff: Mentor

    Since you are asked to find the partial derivative ∂Ψ/∂θ, you can be certain that Ψ is a function of θ and some other variable, most likely r, as the problem is evidently working with polar coordinates. The context of the problem will usually provide information about what variables any functions are in terms of.
    The fact that the flows are shown in an x-y coordinate system is a strong clue.
  6. Dec 16, 2015 #5
    Thank you very much. Makes perfect sense! :)
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