Find the area of the region that is inside the graph of r = 1 and outside of the graph of r = 1 + cos (θ). (4 points)
A) 1.127
B) 1.215
C) 1.275
D) 1.375
To find the area of the region that is inside the graph of r = 1 and outside of the graph of r = 1 + cos(θ), we need to integrate the area element dA over the given region.
The area element in polar coordinates can be expressed as dA = (1/2) r^2 dθ.
First, let's find the limits of integration for the angle θ.
For the region inside the graph of r = 1, θ ranges from 0 to π.
For the region outside the graph of r = 1 + cos(θ), θ ranges from π to 2π.
So, the integral for the area can be written as:
∫[0,π] (1/2)(1)^2 dθ + ∫[π,2π] (1/2)(1+cos(θ))^2 dθ.
Simplifying the integrals, we have:
∫[0,π] (1/2) dθ + ∫[π,2π] (1/2)(1+2cos(θ)+cos^2(θ)) dθ.
Evaluating the integrals, we get:
(1/2)(π-0) + (1/2)(2π - π/2 - 0 + 1/4(π))/2 = (1/2)π + (3π - π/2 + 1/8π).
Combining the terms, we have:
(1/2)π + (3π - π/2 + 1/8π) = (15/8)π - (1/2)π = (7/8)π.
To convert π to decimal form, we divide by π and multiply by 180:
(7/8)π = (7/8)(180) = 157.5 degrees.
Therefore, the area of the region is approximately 157.5 square units.
None of the given options (A), (B), (C), (D) match the calculated value.
To find the area of the region inside the graph of r = 1 and outside the graph of r = 1 + cos(θ), we can use polar coordinates and integrate.
First, let's plot the two polar curves:
- The graph of r = 1 is a circle with a radius of 1 centered at the origin.
- The graph of r = 1 + cos(θ) is another curve that looks like a loop. It represents a cardioid shape.
The area we want to find is the region between these two curves, which lies in the positive quadrant of the coordinate plane. We can focus on the region from θ = 0 to θ = π/2 (the positive x-axis to the positive y-axis).
To find the area, we can set up the integral using the idea of "area between two curves" in polar coordinates. We integrate from θ = 0 to θ = π/2 and subtract the area of the inner curve from the area of the outer curve.
The equation for the area using polar coordinates is:
Area = (1/2) ∫[(outer curve)^2 - (inner curve)^2] dθ
Let's plug in the equations for our curves:
Area = (1/2) ∫[(1 + cos(θ))^2 - 1^2] dθ from θ = 0 to θ = π/2
Simplifying the integrand:
= (1/2) ∫[1 + 2cos(θ) + cos^2(θ) - 1] dθ
= (1/2) ∫[2cos(θ) + cos^2(θ)] dθ
Integrating gives:
Area = (1/2) [2sin(θ) + (θ + (1/2)sin(2θ))] evaluated from θ = 0 to θ = π/2
Plugging in the limits:
Area = (1/2) [2sin(π/2) + (π/2 + (1/2)sin(2(π/2)))] - [(1/2) [2sin(0) + (0 + (1/2)sin(2(0)))]]
Simplifying further:
Area = (1/2) [2 + (π/2 + sin(π))] - (0 + 0)
= (1/2) [2 + (π/2 + 0)] = (1/2)(2 + π/2) = 1 + π/4 ≈ 1.785
So, the area of the region inside the graph of r = 1 and outside the graph of r = 1 + cos(θ) is approximately 1.785.
None of the answer choices provided match this value, so it seems there might be an error in the given options.
The curves intersect where
1+cosθ = 1
θ = ±π/2
But we want the left side of the graphs, so using symmetry
A = 2∫[π/2, 3π/2] 1/2 (1^2 - (1+cosθ)^2) dθ
= ∫[π/2, 3π/2] -2cosθ - cos^2θ dθ
Now finish it off.