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Trouble with Integral to find area using Green's Theorem

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the area of the region within the hypocycloid x[itex]^{2/3}[/itex]+y[itex]^{2/3}[/itex]=a[itex]^{2/3}[/itex] parameterized by x=acos[itex]^{3}[/itex]t, y=asin[itex]^{3}[/itex]t, 0[itex]\leq[/itex]t[itex]\leq[/itex]2[itex]\pi[/itex]

    2. Relevant equations

    In the problem prior to this one, I showed that the line integral of [itex]\vec{F}[/itex]=x[itex]\hat{j}[/itex] around a closed curve in the xy-plane, oriented as in Green's Theorem, measures the area of the region enclosed by the curve. I am supposed to use this fact to find the area above.

    3. The attempt at a solution

    So I've got it all set up right, but I'm having trouble with the integral. Either I'm just forgetting how to solve this sort of integral, or there's another, simpler way to do the problem. The integral I got is:

    [itex]\int[/itex][itex]^{2\pi}_{0}[/itex](3a[itex]^{2}[/itex]sin[itex]^{2}[/itex]tcos[itex]^{4}[/itex]t)dt

    I checked it numerically on wolfram and this integral gives the right answer, but I don't know how to compute it by hand. Normally our prof doesn't assign anything with tricky integrals, so either this is easier than I'm thinking, or there's another way to find the area. But since this vector field isn't path-independent, I don't see any shortcuts I could take.
     
  2. jcsd
  3. Feb 13, 2012 #2

    vela

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    You might try starting by writing everything in terms of sin t.
     
  4. Feb 14, 2012 #3
    Well, I don't remember too many trig identities, but I'll give it a try.

    Starting with sin[itex]^{2}[/itex]t*cos[itex]^{4}[/itex]t (Pulling 3a[itex]^{2}[/itex] out as a constant)

    cos[itex]^{2}[/itex]t = 1-sin[itex]^{2}[/itex]t

    cos[itex]^{4}[/itex]t = (cos[itex]^{2}[/itex]t)[itex]^{2}[/itex] = (1-sin[itex]^{2}[/itex]t)[itex]^{2}[/itex] = 1-2sin[itex]^{2}[/itex]t+sin[itex]^{4}[/itex]t

    sin[itex]^{2}[/itex]t*cos[itex]^{4}[/itex]t = sin[itex]^{2}[/itex]t-2sin[itex]^{4}[/itex]t+sin[itex]^{6}[/itex]t

    Does that look right? The integral still doesn't jump out at me.
     
  5. Feb 14, 2012 #4

    vela

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    That's doable but tedious. Try rewriting the integrand like this:
    $$\sin^2 t\ \cos^4 t = (\sin^2 t\ \cos^2 t)\cos^2 t = \frac{1}{4}\sin^2 2t\left(\frac{1+\cos 2t}{2}\right).$$ That's pretty straightforward to integrate after you multiply it out.
     
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