Find the area of the region which is bounded by the polar curves
theta =pi and
r=2theta 0<theta<1.5pi inclusive
To find the area of the region bounded by the polar curves theta = pi and r = 2theta for 0 < theta < 1.5pi (inclusive), we can use the concept of integration.
The first step is to visualize the region by plotting the two polar curves on a graph. The curve theta = pi is a vertical line passing through the point (-pi, r). The curve r = 2theta represents a spiral extending outward from the origin.
To find the area, we need to determine the limits of integration for both r and theta. The given conditions state that the angle theta varies from 0 to 1.5pi, which includes the interval [0, 1.5pi]. The radius r = 2theta depends on theta.
Next, we need to set up the integral in terms of the polar coordinates. The area of a small, infinitesimal element in polar coordinates is dA = 0.5 * r^2 * d(theta). We can express r in terms of theta, which gives us r = 2theta.
Now, we can set up the integral for the area of the region:
A = ∫(theta = 0 to 1.5pi) ∫(r = 0 to 2theta) 0.5 * r^2 d(theta)
Simplifying the integral:
A = 0.5 * ∫(theta = 0 to 1.5pi) ∫(r = 0 to 2theta) (2theta)^2 d(theta)
A = 0.5 * ∫(theta = 0 to 1.5pi) ∫(r = 0 to 2theta) 4(theta^2) d(theta)
Integrating with respect to r and theta:
A = 0.5 * ∫(theta = 0 to 1.5pi) [4/3 * (2theta)^3] |(r = 0 to 2theta) d(theta)
A = 0.5 * ∫(theta = 0 to 1.5pi) 4/3 * (8theta^3) d(theta)
Simplifying further:
A = 0.5 * (32/3) * ∫(theta = 0 to 1.5pi) theta^3 d(theta)
A = (16/3) * ∫(theta = 0 to 1.5pi) theta^3 d(theta)
To find the antiderivative of theta^3, we can use the power rule for integration:
A = (16/3) * [theta^4 / 4] |(theta = 0 to 1.5pi)
A = (16/3) * [(1.5pi)^4 / 4 - 0^4 / 4]
A = (16/3) * (5.0625pi^4)
Thus, the area of the region bounded by the polar curves theta = pi and r = 2theta for 0 < theta < 1.5pi (inclusive) is (16/3) * (5.0625pi^4).