Find the area of the region which is bounded by the polar curves
theta =pi and
r=2theta 0<theta<1.5pi inclusive
The elemental area of an isosceles triangle subtending an angle of δθ and radius(length) r is
(1/2)r²δθ.
The required area is therefore:
∫(1/2)r²δθ
for θ between the given limits (0 orπ?) to 1.5π.
You will need to substitute r by 2θ as in the given function.
As I noted in my original answer, there is an inconsistency in the specified limits of integration or "boundaries" of your enclosed area.
The spiral curve r = 2 theta starts at theta = 0 and ends at theta = 1.5 pi. The straight line theta = pi is the -x axis. It is not clear how the area from theta = pi to theta = 3/2 pi should be treated when doing the area integration.
You have been told how to do the integration. Try various integration limits (0 to pi; 0 to 3/2 pi; pi to 3/2 pi) and see which your grading software accepts.
To find the area of the region bounded by the polar curves, we need to integrate the formula for the area bounded by two polar curves.
The formula for the area bounded by two polar curves, r = f(θ) and r = g(θ), where θ lies between α and β, is given by the integral ∫[α,β] 1/2 (f(θ))^2 - 1/2 (g(θ))^2 dθ.
In this case, the polar curves given are:
θ = π (a vertical line passing through the origin)
r = 2θ (a spiral)
To find the area bounded by these curves, we need to determine the values of θ where they intersect. By setting the equations equal to each other, we have:
2θ = π
Solving for θ, we get:
θ = π/2
So, the region of interest lies between θ = π/2 and θ = 1.5π.
Now, we can calculate the area using the formula mentioned above:
∫[π/2,1.5π] 1/2 (2θ)^2 - 1/2 (0)^2 dθ
Simplifying the integral, we have:
∫[π/2, 1.5π] 2θ^2 dθ
Integrating with respect to θ, we get:
(2/3)θ^3 | [π/2, 1.5π]
Evaluating the integral at the limits, we have:
(2/3)(1.5π)^3 - (2/3)(π/2)^3
Simplifying further, we get:
(9/8π^3) - (1/8π^3) = (8/8π^3) = 1/π^3
Therefore, the area of the region bounded by the polar curves θ = π and r = 2θ, where 0 < θ < 1.5π inclusive, is 1/π^3 square units.