# Integral of polar curve

1. Apr 16, 2014

### syeh

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

to find are under a curve you find integral under the curve

3. The attempt at a solution
the solutions are:
a) 2*(1/2) ∫(0 to pi) (2-cosΘ)2

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???

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2. Apr 16, 2014

### SammyS

Staff Emeritus

Do you really understand as the above statement claims? It almost answers the following.

Except that taking the limits of integration from 0 to π is what gives a semi-circle rather than multiplying by 1/2 .

3. Apr 16, 2014

### BiGyElLoWhAt

it looks like to me they (for a)they just simplified.

What you want is the area here: r=2-cosΘ for 0≤θ≤2pi

What you start out with is: $∫_{E}dA$ Then you begin to make substitutions. and you're right, you can't do a standard integral, so you have to break your region (E) up into to parts, from 0 to pi, and from pi to 2 pi, so now we have $2∫_{0}^{\pi}dA$, but what is dA? In polar it's rdrdΘ.

Now what we have is $2∫_{0}^{\pi}∫_{0}^{2-cos(\theta )}rdrd\theta$
This gives you $2∫_{0}^{pi}\frac{r^2}{2}d\theta|_{r=0}^{r=2-cos(\theta)}$

evaluate it and theres a)

4. Apr 16, 2014

### haruspex

I don't think that's what they did. Consider a triangle subtended at the origin by the segment of the curve starting at (r, θ) and length rdθ. What is its area?