Consider the 4 leaf rose and a circle having polar equations: r=10cos(2Θ) and r=5 with 0≤Θ≤2π, respectively. Find the area of the region that lies inside the rose and outside the circle. hint: find the smallest positive value of Θ for which the two curves intersect; use symmetry.
So far I have gotten; 1/2*int[-π/4,π/4](10cos^2(2Θ))dΘ
=5/2*int[-π/4,π/4](1+cos(4Θ))dΘ
I am not sure about my lower and upper limits. Will this set up get me the answer?
The curves intersect at π/6, not π/4. The leaves of the rosette return to (0,0) at multiples of π/4, but that's not where the circle intersects the rosette.
Using the 8-way symmetry, I'd just integrate over [0,π/6], so
a = 8∫[0,π/6] 1/2 (R^2-r^2) dθ
where R=10cos(2θ) and r=5, so
a = 8∫[0,π/6] 1/2 (100cos^2(2θ)-25) dθ = 25/3 (3√3+2π)
Your setup looks correct so far. However, the limits of integration need to be adjusted to take into account the symmetry of the problem.
Since the polar equations have a symmetric shape, you can find the area of one petal and then multiply it by the number of petals to get the total area. In this case, there are four petals, so you need to integrate over only one-fourth of the domain.
To find the limits of integration, you need to determine where the two curves intersect. This occurs when the values of "r" for the rose equation and the circle equation are equal.
For the rose equation, we have r = 10cos(2Θ).
For the circle equation, we have r = 5.
Setting them equal, we can solve for Θ:
10cos(2Θ) = 5
cos(2Θ) = 0.5
2Θ = π/3 or 2Θ = 5π/3
Θ = π/6 or Θ = 5π/6
Since the rose curve is symmetric, we only need to consider the positive values of Θ.
The limits of integration for the area of one petal should be from Θ = 0 to Θ = π/6. However, your current limits of integration are from Θ = -π/4 to Θ = π/4, which are not correct.
To get the correct setup, you should integrate over the interval [0, π/6] instead of [-π/4, π/4]. Using this revised setup, your calculation will yield the area of one petal.
Finally, to obtain the total area including all four petals, multiply this result by four.