Use iterated integral to find the area enclosed by r=2sin2(theta).
How do i graph this?
For the graph, see
http://www.wolframalpha.com/input/?i=+r+%3D+2sin%5E2%CE%B8
For the area, recall that in polar coordinates,
a = ∫∫ r dr dθ
= ∫[0,2π] ∫[0,2sin^2θ] r dr dθ
= ∫[0,2π] 1/2 r^2 dθ [[0,2sin^2θ]
= ∫[0,2π] 1/2 (2sin^2θ)^2 dθ
= ∫[0,2π] 2 sin^4θ dθ
= 3π/2
You have to use integration by parts twice, so I'll leave the details to you.
To graph the equation r = 2sin^2(theta), you can follow these steps:
1. Set up a polar coordinate system: Draw a horizontal x-axis (polar axis) and a vertical y-axis (polar axis) that intersect at the origin.
2. Determine the appropriate range for theta: Since the equation involves sin^2(theta), which ranges from 0 to 1, you should choose the interval for theta accordingly. In this case, an appropriate range is from 0 to 2pi, which represents one complete revolution.
3. Plot points: For various values of theta in the chosen interval, compute r by substituting theta into the equation. Then, plot each point (r, theta) on the polar coordinate system to obtain a set of points.
4. Connect the points: Once you have plotted enough points, connect them smoothly to create the curve. This curve represents the graph of the polar equation r = 2sin^2(theta).
By following these steps, you should be able to graph the equation r = 2sin^2(theta) and visualize the enclosed area.