Find the area of the region which is bounded by the polar curves
theta =pi and
r=2theta 0<theta<1.5pi inclusive
It will help to draw a figure.
I did, and it looks to me like the line theta = pi is NOT a bounding curve. The enclosed area is between the theta = 1.5 line and the spiral cruve r = 2 theta.
For the area, integrate
(1/2) r^2 d theta
= (1/2) 4 theta^2 dtheta
from theta = 0 to theta = 1.5 pi
I get the indefinite interal to be (2/3)(theta)^3
The definite integral is then
(2/3)(3 pi/2)^3
= (9/4) pi^3
The answer is not correct, please try again.
I got it, u have to
integrate (1/2) 4 theta^2 dtheta
from theta=0 to theta=pi
To find the area of the region bounded by polar curves, we need to integrate the function representing the curves over the specified range of theta. Here are the steps to find the area:
1. Convert the equations of the polar curves into Cartesian form.
2. Determine the range of theta values for which the curves overlap.
3. Set up the integral to find the area using the formula: A = ∫[lower_bound, upper_bound] (1/2) * r^2 * d(theta).
4. Evaluate the integral to get the final result.
Let's go through each step in detail:
1. Convert the polar equations into Cartesian form:
- The equation for the curve theta = pi represents a vertical line passing through the point (x, y) = (-1, 0).
- The equation for the curve r = 2theta can be expressed in Cartesian coordinates as x^2 + y^2 = 2(theta), keeping in mind that r = sqrt(x^2 + y^2).
2. Determine the range of theta values:
The given range is 0 ≤ theta ≤ 1.5pi, inclusive. This means we need to evaluate the area within this theta range.
3. Set up the integral:
The formula for the area within a polar region is A = ∫(1/2) * r^2 * d(theta), where r is the distance from the origin to a point on the curve.
In this case, the integral becomes A = ∫(1/2) * (x^2 + y^2) * d(theta), where x^2 + y^2 = 2(theta) and -1 ≤ x ≤ 0.
A = ∫(1/2) * (2(theta)) * d(theta) over the range 0 ≤ theta ≤ 1.5pi.
4. Evaluate the integral:
Integrating ∫(1/2) * (2(theta)) * d(theta) with the appropriate bounds will give us the area of the region.
Let's evaluate the integral:
A = ∫(1/2) * (2(theta)) * d(theta) from 0 to 1.5pi
A = [θ^2] evaluated from 0 to 1.5pi
A = (1.5pi)^2 - (0)^2
A = 2.25pi^2
Therefore, the area of the region bounded by the given polar curves is 2.25pi^2 square units.