Question

Find the area of the region shared by the circles r=1, and r=2sin(theta)

Answer 1

Full area = 4*area = 5.5584

Answer 2

I need help getting the problem started.

Answer 3

r = 1 is a circle centered at the origin
r = 2sin(theta) is a figure 8 oriented along the y-axis

You just need to find the area in one quadrant and then multiply by 4 to get the total area.

I am going to use:
T = theta
r1 = 2sin(T) ..... and use A1 to represent area inside this curve
r2 = 1 ............. and use A2 to represent area inside this curve

First find point of intersection:
1 = 2sin(T)
sin(T) = (1/2) and T = 30 degress (pi/6 radians).

So there are two regions in which you need the area. One from 0 to pi/6 and the other from pi6/ to pi/2. In the first r1 will be the outer edge and in the second it will be r2.

The area from pi/6 to pi/2 will be divided into two regions. One will be the area inside r1 and the other the area inside r2. The area we want will then be the difference of these two:
Area = A1 - A2

dA1 = dr(rdT) = rdrdT
[r^2/2]dT
[(2sin(T)^2/2]dT = 2sin^2(T)dT
2[T/2 - sin(2T)/4] = T - sin(2T)/2
A1 =[pi/2] - [pi/12 - SQRT(3)/4] = 5pi/12 + SQRT(3)/4
A1 = 1.309

A2 is just part of a circle and it is the section between theta = pi/6 and pi/2 which is pi/3 or 1/6 of the circle. the area is then piR^2/6 = pi/6

Area1 = A1 - A2 = 5pi/12 + SQRT(3)/4 - pi/6
Area1 = A1 - A2 = pi/4 + SQRT(3/4) = 1.2184

The area from 0 to pi/6 will also be treated in the same way. Only this tiem the area will be A2 - A1.

Again A2 is just a section of a circle and in this case that between 0 and pi/6 or 1/12 of a circle. So A2 is pi/12.

Integrate to get the other area.
the element of area is rdrdT
integrate over r to get: [r^2/2]dT evaluated from 0 to 2sin(T)

2sin^2(T)dT
Integrate this over T:
2[T/2 + sin(2T)/4] evaluated from 0 to pi/6
T - sin(2T)/2
A2 = [pi/6 + (SQRT(3)/4] - [0]

Area2 = A1 - A2 = pi/12 - [pi/6 - SQRT(3)/4]
Area2 = SQRT(3)/4 - pi/12 = 0.1712

Total area of in first quadrant = Area1 + Area2 = 1.2184 + 0.1712
area = 1.3896

Answer 4

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