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Trouble with Integral to find area using Green's Theorem

  • Thread starter Opus_723
  • Start date
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.
 

vela

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You might try starting by writing everything in terms of sin t.
 
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.
 

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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