Integral of Polar Curve

[syeh]

Homework Statement

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.

Homework Equations

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 don't understand why they took (2-cosθ) and squared it...? also why the limits of integration are 0 to pi, not 0 to 2pi?

 

[SammyS]

Homework Statement

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.

Homework Equations

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 don't understand why they took (2-cosθ) and squared it...? also why the limits of integration are 0 to pi, not 0 to 2pi?

You have answered your own final question.

...

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.
...

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

... also why the limits of integration are 0 to pi, not 0 to 2pi?

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

 

[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 there's a)

 

[haruspex]

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 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?