find the area of the region bounded by the polar curve r=sqrt(6ln(theta)+3
as well as the rays theta=1 and theta=e
To find the area of the region bounded by the polar curve r = √(6ln(θ) + 3) and the rays θ = 1 and θ = e, we need to integrate the area between the curve and the x-axis.
First, let's solve for the intersection points of the curve with the rays.
For the ray θ = 1:
We substitute θ = 1 into r = √(6ln(θ) + 3) to find the corresponding value of r:
r = √(6ln(1) + 3) = √(6 * 0 + 3) = √3
So, at θ = 1, r = √3.
For the ray θ = e:
We substitute θ = e into r = √(6ln(θ) + 3) to find the corresponding value of r:
r = √(6ln(e) + 3) = √(6 * 1 + 3) = √9 = 3
So, at θ = e, r = 3.
To set up the integral for the area, we integrate the function πr^2 with respect to θ, between θ = 1 and θ = e:
Area = ∫[θ=1 to θ=e] πr^2 dθ
Area = π ∫[θ=1 to θ=e] (6ln(θ) + 3) dθ
Next, we calculate the integral:
∫ (6ln(θ) + 3) dθ = 6 ∫ln(θ) dθ + 3 ∫1 dθ
The integral of ln(θ) can be found by using integration by parts:
∫ln(θ) dθ = θln(θ) - ∫(1/θ)θ dθ
= θln(θ) - ∫(1) dθ
= θln(θ) - θ + C
So, the integral becomes:
6 ∫ln(θ) dθ + 3 ∫1 dθ = 6(θln(θ) - θ) + 3θ + C
Now, we substitute the limits of integration:
Area = π [6(θln(θ) - θ) + 3θ] |[θ=1 to θ=e]
Area = π[6(e ln(e) - e) + 3e - 6(1 ln(1) - 1) - 3]
Since ln(1) = 0, we can simplify further:
Area = π[6e - 6e + 3e - 6 + 3]
Area = π[3e - 3]
So, the area of the region bounded by the polar curve r = √(6ln(θ) + 3) and the rays θ = 1 and θ = e is π(3e - 3).
To find the area of the region bounded by the polar curve r = √(6ln(θ) + 3) and the rays θ = 1 and θ = e, you can use the formula for the area in polar coordinates.
The formula to find the area of a region bounded by a polar curve is:
A = (1/2) ∫[θ1,θ2] (r(θ))^2 dθ
In this case, θ1 = 1 and θ2 = e, and r(θ) = √(6ln(θ) + 3).
So, the area A can be calculated as:
A = (1/2) ∫[1,e] (√(6ln(θ) + 3))^2 dθ
To evaluate this integral, you can simplify the integrand:
(√(6ln(θ) + 3))^2 = (6ln(θ) + 3)
Now you can integrate:
A = (1/2) ∫[1,e] (6ln(θ) + 3) dθ
Integrating term by term:
A = (1/2) [6∫[1,e] ln(θ) dθ + 3∫[1,e] dθ]
The integral of ln(θ) is θln(θ) - θ + C1, where C1 is the constant of integration.
The integral of dθ is θ + C2, where C2 is the constant of integration.
So, the area becomes:
A = (1/2) [6(θln(θ) - θ)|[1,e] + 3(θ)|[1,e]]
Evaluating the integral limits:
A = (1/2) [6(e ln(e) - e - (1ln(1) - 1)) + 3(e - 1)]
Simplifying:
A = (1/2) [6e - 6 - 3 + 3e - 3]
A = (1/2) [9e - 12]
Therefore, the area of the region bounded by the polar curve r = √(6ln(θ) + 3), θ = 1, and θ = e is (9e - 12)/2 square units.