Integration of part of a circle

[Sheve]

Homework Statement

Draw a circle centered on the origin with radius 2. Draw the vertical line x = 1. Find the area bounded on the left by x=1 and on the right by the circle.

Homework Equations

Area of a circle: x^{2} + y^{2} = r^{2}. In this race, r = 2

Polar coordinates:

x = p cos(\theta)
y = p sin(\theta)

The Attempt at a Solution

At first I tried to solve this in rectangular coordinates, using the equation y = \sqrt{4 - x^{2}} and integrating it from 1 to 2. However the antiderivative for that is messy as heck, and this is supposed to be a simple problem. I'm trying to do it in polar coords, but I'm at a loss as how to bound the integral in that case.

 

[Mark44]

In the region of concern, r ranges from r = 1/cos(theta) = sec(theta) to r = 2. Theta ranges from theta = -arctan(sqrt(3)) to theta = +arctan(sqrt(3)), but you can let theta run from 0 to arctan(sqrt(3)) and double the value, using the symmetry of the circle.

The integral in rectangular coordinates isn't that bad if you know trig substitution.

 

[Sheve]

I was hoping to avoid having to use arctan if at all possible...I'm having more luck with the small triangles method in polar coordinates.

You take a triangle with height along the x axis, "r". The angle from the x-axis is assumed to be very small, d\theta. Because this is very small, the base of the triangle is approximated as the arc length, given in this case by r * d\theta. The area of a triangle is 1/2 * base * height, or in this case 1/2*r^{2}d\theta. Integrating this from 0 to 2pi gives you your answer.

 

[Mark44]

I was hoping to avoid having to use arctan if at all possible...

Arctan doesn't enter into the picture at all. If you stay with rectangular coordinates, the substitution is sin(theta) = x/4

I'm having more luck with the small triangles method in polar coordinates.

You take a triangle with height along the x axis, "r".

You lost me here. How can the height be along the x axis? And what does this have to do with r?

The angle from the x-axis is assumed to be very small, d\theta. Because this is very small, the base of the triangle is approximated as the arc length, given in this case by r * d\theta. The area of a triangle is 1/2 * base * height, or in this case 1/2*r^{2}d\theta. Integrating this from 0 to 2pi gives you your answer.

Why would you integrate from 0 to 2pi? In polar coordinates, the line x = 1 intersects the upper part of the circle at (2, pi/3).