Calculus 2 Polar Coordinate System

Question

A region is bounded by the line y = x, y = x√3(radical 3), y = 1/2 and the circle x^2 + (y-1)^2 = 1. Draw the region and set up the integral in polar coordinate system that represents the area of the region. Do not evaluate the integral.

Answer 1

I can't do picture in this editor. But integral for area A = 1/2∫_π/4^π/3(4sin²θ - 1/(2sinθ)²)dθ

Answer 2

I've drawn the region here: https://imgur.com/ENh7EyI

The red line is y = x, the blue line is y = x√3, the orange line is x² + (y-1)² = 1, and the green line is y = 1/2.

To find the area of the black region using polar coordinates, we have to find the bounds of the region. We can think of sweeping out this area by rotating a ray extending from the origin counterclockwise from the line y = x to the line y = x√3. That is exactly what we will be varying: the angle θ in our polar coordinate frame. At each particular angle, we sweep out some small slice of the region which has a maximum radius along the orange circle and a minimum radius along the green line y = 1/2.

But if we're to find the area using this idea, we need to convert these bounds into the language of polar coordinates. As noted, the bounds on θ will be given by the red and the blue lines. The red line y = x has a slope of 1 and so is at an angle of arctan(1) = π/4 from the positive x-axis. Meanwhile, the blue line y = x√3 has a slope of √3 and so is at an angle of arctan(√3) = π/3 from the positive x-axis. Thus, we will be integrating over π/4 ≤ θ ≤ π/3.

The maximum radius in any given slice is given by the distance to the origin of the orange circle x² + (y-1)² = 1. To convert this into polar coordinates, use the fact that x = r*cos(θ) and y = r*sin(θ) and turn the equation into a function r(θ):

1. x² + (y-1)² = 1 ⇒ (r*cos(θ))² + (r*sin(θ)-1)² = 1
2. ⇒ r²cos²(θ) + r²sin²(θ) - 2r*sin(θ) + 1 = 1
3. ⇒ r²(cos²(θ) + sin²(θ)) = 2r*sin(θ)
4. ⇒ r = 2*sin(θ)

We can use the same method to convert the minimum radius of each slice, given by the green line y = 1/2, into polar coordinates:

1. y = 1/2 ⇒ r*sin(θ) = 1/2
2. ⇒ r = 1/2 * csc(θ)

Now we are ready to integrate. The area formula for polar coordinates is:

1. A = ∫_{α ≤ θ ≤ β} (1/2)*(R² - r²) dθ

So we can just plug the values in:

1. A = ∫_{π/4 ≤ θ ≤ π/3} (1/2)*((2*sin(θ))² - (1/2 * csc(θ))²) dθ

And we are done, since we are not supposed to evaluate the integral.