Question
Find the area inside the larger loop and outside the smaller loop of the curve.
r = (1/2) + sin(theta)
r = (1/2) + sin(theta)
Answers
Answered by
oobleck
This statement is clearly false:
the smaller loop lies between 0 and π and the larger loop lies between π and 2π
if you look at the graph.
the smaller loop lies between 0 and π and the larger loop lies between π and 2π
if you look at the graph.
Answered by
oobleck
Actually, since the smaller loop lies inside the larger loop, its area must be subtracted to find the portion outside the smaller loop.
That makes the area
∫[-π/6,7π/6] 1/2 (1/2 + sinθ)^2 dθ - ∫[7π/6,11π/6] 1/2 (1/2 + sinθ)^2 dθ
= 1/8 (4π+3√3) - 1/8 (2π+3√3)
= 1/4 (π+3√3)
That makes the area
∫[-π/6,7π/6] 1/2 (1/2 + sinθ)^2 dθ - ∫[7π/6,11π/6] 1/2 (1/2 + sinθ)^2 dθ
= 1/8 (4π+3√3) - 1/8 (2π+3√3)
= 1/4 (π+3√3)
Answered by
GPT 3.5
To find the area inside the larger loop and outside the smaller loop of the curve, we need to determine the limits of integration for the given polar equation.
First, let's graph the given curve r = (1/2) + sin(theta) to visualize the loops:
Graph of r = (1/2) + sin(theta)
As we can see from the graph, there are two loops - one smaller and one larger.
To find the area inside the larger loop and outside the smaller loop, we need to find the values of theta that define the loops and set up the integral accordingly.
To determine the limits of integration, we need to find the values of theta where r = (1/2) + sin(theta) crosses the x-axis.
When r = 0, we have:
0 = (1/2) + sin(theta)
sin(theta) = -1/2
Using the unit circle, we see that sin(theta) = -1/2 corresponds to theta = 7π/6 and 11π/6.
Since the smaller loop lies between 0 and π and the larger loop lies between π and 2π, the limits of integration for the area between the loops are θ = π to 2π.
The area inside the larger loop and outside the smaller loop can be found using the formula for the area in polar coordinates:
A = 1/2 ∫[θ = π to 2π] r^2 dθ
Substituting the given equation r = (1/2) + sin(theta) into the formula, we have:
A = 1/2 ∫[θ = π to 2π] [(1/2) + sin(theta)]^2 dθ
Simplifying and expanding the square, we get:
A = 1/2 ∫[θ = π to 2π] [(1/4) + (1/2)sin(theta) + sin^2(theta)] dθ
Next, we integrate term by term:
A = 1/2 [(1/4)θ + (1/2)(-cos(theta)) + (1/2)θ - (1/4)sin(2θ)] evaluated from θ = π to 2π
A = 1/2 [(2π - π)/4 + (1/2)(-cos(2π) - (-cos(π))) + (2π - π)/2 - (1/4)(sin(4π) - sin(2π))]
Simplifying further:
A = 1/2 [π/4 - 1/2 + π/2 - 1/4(sin(4π) - sin(2π))]
A = 1/2 [π/4 - 1/2 + π/2 - 1/4(0 - 0)]
A = 1/2 [π/4 - 1/2 + π/2]
A = 1/2 [π/4 + π - 1/2]
A = 1/2 (3π/4 - 1/2)
Therefore, the area inside the larger loop and outside the smaller loop of the curve r = (1/2) + sin(theta) is (3π/8 - 1/4) square units.
First, let's graph the given curve r = (1/2) + sin(theta) to visualize the loops:
Graph of r = (1/2) + sin(theta)
As we can see from the graph, there are two loops - one smaller and one larger.
To find the area inside the larger loop and outside the smaller loop, we need to find the values of theta that define the loops and set up the integral accordingly.
To determine the limits of integration, we need to find the values of theta where r = (1/2) + sin(theta) crosses the x-axis.
When r = 0, we have:
0 = (1/2) + sin(theta)
sin(theta) = -1/2
Using the unit circle, we see that sin(theta) = -1/2 corresponds to theta = 7π/6 and 11π/6.
Since the smaller loop lies between 0 and π and the larger loop lies between π and 2π, the limits of integration for the area between the loops are θ = π to 2π.
The area inside the larger loop and outside the smaller loop can be found using the formula for the area in polar coordinates:
A = 1/2 ∫[θ = π to 2π] r^2 dθ
Substituting the given equation r = (1/2) + sin(theta) into the formula, we have:
A = 1/2 ∫[θ = π to 2π] [(1/2) + sin(theta)]^2 dθ
Simplifying and expanding the square, we get:
A = 1/2 ∫[θ = π to 2π] [(1/4) + (1/2)sin(theta) + sin^2(theta)] dθ
Next, we integrate term by term:
A = 1/2 [(1/4)θ + (1/2)(-cos(theta)) + (1/2)θ - (1/4)sin(2θ)] evaluated from θ = π to 2π
A = 1/2 [(2π - π)/4 + (1/2)(-cos(2π) - (-cos(π))) + (2π - π)/2 - (1/4)(sin(4π) - sin(2π))]
Simplifying further:
A = 1/2 [π/4 - 1/2 + π/2 - 1/4(sin(4π) - sin(2π))]
A = 1/2 [π/4 - 1/2 + π/2 - 1/4(0 - 0)]
A = 1/2 [π/4 - 1/2 + π/2]
A = 1/2 [π/4 + π - 1/2]
A = 1/2 (3π/4 - 1/2)
Therefore, the area inside the larger loop and outside the smaller loop of the curve r = (1/2) + sin(theta) is (3π/8 - 1/4) square units.
Answered by
GPT 3.5
You are correct, I apologize for the mistake in my previous response. The smaller loop of the curve lies between θ = 7π/6 and 11π/6, while the larger loop lies between θ = π/6 and 13π/6.
To find the area inside the larger loop and outside the smaller loop, we need to set up the integral accordingly. The area can be calculated using the formula for the area in polar coordinates:
A = 1/2 ∫[θ = π/6 to 13π/6] r^2 dθ
Substituting the given equation r = (1/2) + sin(theta) into the formula, we have:
A = 1/2 ∫[θ = π/6 to 13π/6] [(1/2) + sin(theta)]^2 dθ
Expanding the square:
A = 1/2 ∫[θ = π/6 to 13π/6] [(1/4) + sin(theta) + sin^2(theta)] dθ
Integrating term by term:
A = 1/2 [(1/4)θ - cos(theta) - (1/2) cos^2(theta)] evaluated from θ = π/6 to 13π/6
Simplifying further:
A = 1/2 [(1/4)(13π/6 - π/6) - cos(13π/6) - (1/2)cos^2(13π/6) - ((1/4)(π/6 - π/6) - cos(π/6) - (1/2)cos^2(π/6))]
A = 1/2 [(12π/24) - cos(13π/6) - (1/2)cos^2(13π/6) - (0 - cos(π/6) - (1/2)cos^2(π/6))]
A = 1/2 [(π/2) - cos(13π/6) - (1/2)cos^2(13π/6) - (0 - sqrt(3)/2 - (1/2)(3/4))]
A = 1/2 [(π/2) - cos(13π/6) - (1/8)(3 + cos(13π/6))]
Simplifying further:
A = 1/2 [(8π - 13π)/24 - (8 + cos(13π/6))/16]
A = 1/2 [(π - 13π)/24 - (8 + 1)/16]
A = 1/2 [(π/24 - 13π/24) - 9/16]
A = 1/2 [-12π/24 - 9/16]
A = 1/2 [-π/2 - 9/16]
Therefore, the area inside the larger loop and outside the smaller loop of the curve r = (1/2) + sin(theta) is (-π/4 - 9/32) square units.
To find the area inside the larger loop and outside the smaller loop, we need to set up the integral accordingly. The area can be calculated using the formula for the area in polar coordinates:
A = 1/2 ∫[θ = π/6 to 13π/6] r^2 dθ
Substituting the given equation r = (1/2) + sin(theta) into the formula, we have:
A = 1/2 ∫[θ = π/6 to 13π/6] [(1/2) + sin(theta)]^2 dθ
Expanding the square:
A = 1/2 ∫[θ = π/6 to 13π/6] [(1/4) + sin(theta) + sin^2(theta)] dθ
Integrating term by term:
A = 1/2 [(1/4)θ - cos(theta) - (1/2) cos^2(theta)] evaluated from θ = π/6 to 13π/6
Simplifying further:
A = 1/2 [(1/4)(13π/6 - π/6) - cos(13π/6) - (1/2)cos^2(13π/6) - ((1/4)(π/6 - π/6) - cos(π/6) - (1/2)cos^2(π/6))]
A = 1/2 [(12π/24) - cos(13π/6) - (1/2)cos^2(13π/6) - (0 - cos(π/6) - (1/2)cos^2(π/6))]
A = 1/2 [(π/2) - cos(13π/6) - (1/2)cos^2(13π/6) - (0 - sqrt(3)/2 - (1/2)(3/4))]
A = 1/2 [(π/2) - cos(13π/6) - (1/8)(3 + cos(13π/6))]
Simplifying further:
A = 1/2 [(8π - 13π)/24 - (8 + cos(13π/6))/16]
A = 1/2 [(π - 13π)/24 - (8 + 1)/16]
A = 1/2 [(π/24 - 13π/24) - 9/16]
A = 1/2 [-12π/24 - 9/16]
A = 1/2 [-π/2 - 9/16]
Therefore, the area inside the larger loop and outside the smaller loop of the curve r = (1/2) + sin(theta) is (-π/4 - 9/32) square units.
Answered by
GPT 3.5
You are correct, I apologize for the mistake in my previous response. Thank you for catching that error. The correct area inside the larger loop and outside the smaller loop of the curve r = (1/2) + sin(theta) is indeed 1/4 (π+3√3) square units. Thank you for pointing out the mistake.