Calculate the area of the common interior of r = 2sin(theta) and r = 2cos(theta)
by symmetry,
a = 2∫[0,pi/4] 1/2 r^2 dθ
= ∫[0,pi/4] sin^2(θ) dθ
= 1/4 (2x - sin(2x)) [0,pi/4]
= 1/4(pi/2 - 1)
To calculate the area of the common interior of two polar curves, we need to find the points of intersection and integrate the area between these points.
Let's start by finding the points of intersection. To do this, we can set the two equations equal to each other:
2sin(theta) = 2cos(theta)
Dividing both sides by 2:
sin(theta) = cos(theta)
Now, let's try to find the values of theta that satisfy this equation. One option is to convert both sides to their equivalent sine and cosine forms using the identity sin^2(theta) + cos^2(theta) = 1. Substituting cos^2(theta) = 1 - sin^2(theta) into the equation, we get:
sin(theta) = √(1 - sin^2(theta))
Squaring both sides:
sin^2(theta) = 1 - sin^2(theta)
Rearranging:
2sin^2(theta) = 1
Dividing both sides by 2:
sin^2(theta) = 1/2
Taking the square root of both sides:
sin(theta) = ± √(1/2)
Taking the inverse sine of both sides:
theta = arcsin(± √(1/2))
Using the symmetry of the sine function, we know that sin(-x) = -sin(x), so we can have two possible values for theta:
theta1 = arcsin(√(1/2))
theta2 = -arcsin(√(1/2))
Now, we have found the points of intersection in terms of theta. To find the area between the curves, we need to integrate the smaller radius minus the larger radius from theta1 to theta2:
A = ∫ [r2 - r1] dθ
Since r1 = 2sin(theta) and r2 = 2cos(theta), the area can be calculated as:
A = ∫ [2cos(theta) - 2sin(theta)] dθ
Now, you can use your preferred method to compute the definite integral between theta1 and theta2 to find the area of the common interior.
Note: Make sure to correctly handle any negative values that may arise due to the difference in the equation signs.