Express the region inside the leaf of the rose r=2sin2? in the first quadrant as an iterated integral over R?
Something is wrong with your sine function, see this.
http://answers.yahoo.com/question/index?qid=20100428182120AAzOqL1
yeah I forgot to type in the theta
To express the region inside the leaf of the rose r = 2sin(2θ) in the first quadrant as an iterated integral over R, we first need to determine the limits of integration.
In polar coordinates, the first quadrant corresponds to θ ranging from 0 to π/2, and the radial distance r can vary between 0 and the value of the rose equation at a given θ.
Let's solve for the intersection points of the rose with the x-axis (r = 0) in order to find the limits of integration for r. Set r = 0 in the equation r = 2sin(2θ):
0 = 2sin(2θ)
sin(2θ) = 0
The equation sin(2θ) = 0 is satisfied when 2θ is a multiple of π (0, π, 2π, etc.). Therefore, when θ = 0, π, and 2π, the rose intersects the x-axis.
Since we are only interested in the first quadrant, the limits of integration for θ become 0 to π/2. The limits of integration for r are from 0 to the value of the rose equation at a given θ, which is 2sin(2θ).
Now, we can express the region as an iterated integral over R:
∫∫R dA = ∫₀^(π/2) ∫₀^(2sin(2θ)) r dr dθ
In this integral, the outer integral is with respect to θ, and the inner integral is with respect to r.
Note: The region R is the region bounded by the leaf of the rose in the first quadrant.