The graph of the polar curve r=2-cosΘ for 0≤θ≤2pi is shown in the figure. (attached)
a) write an integral expression for the area of the region inside the curve
b) write expressions for dx/dΘ and dy/dΘ in terms of Θ
c) find dy/dx as a function of Θ
d) write an equation in terms of x and y for the line tangent to the graph of the polar curve at the point where Θ=pi/2. show the work that lead to your answer.
to find are under a curve you find integral under the curve
The Attempt at a Solution
the solutions are:
a) 2*(1/2) ∫(0 to pi) (2-cosΘ)2dΘ
b) dx/dΘ = -2sinΘ + 2(cosΘ)(sinΘ)
dy/dΘ = 2cosΘ-cos2Θ + sin2Θ
c) dy/dx = (2cosΘ-cos2Θ +sin2θ) / (-2sinθ + 2(cosθ)(sinθ))
So, i tried part A but could not figure out the area of the region because if you did a regular integral of ∫(2-cosθ), according to the graph the region above and below the x-axis would cancel out, resulting in an answer of 0.
i understand that you might have to change the equation into a semi-circle by multiplying 1/2 then later have to undo that by multiplying by 2.
i dont understand why they took (2-cosθ) and squared it...? also why the limits of integration are 0 to pi, not 0 to 2pi???