Find the area inside the larger loop and outside the smaller loop of the curve.
r = (1/2) + sin(theta)
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.
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.
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)
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.
To find the area inside the larger loop and outside the smaller loop of the curve given by r = (1/2) + sin(theta), we'll need to use the concept of polar coordinates.
Step 1: Determine the points of intersection between the two curves.
To find the points of intersection, we need to equate the two equations for r:
(1/2) + sin(theta) = r
(1/2) + sin(theta) = (1/2) + sin(theta)
This equation holds for all values of theta. Therefore, the two curves r = (1/2) + sin(theta) intersect at all points in the polar coordinate system.
Step 2: Determine the limits of integration for theta.
To find the area inside the larger loop and outside the smaller loop, we'll integrate with respect to theta. The limits of integration for theta will be the values where the curves intersect.
Since the curves intersect for all values of theta, the limits of integration for theta are 0 to 2π. Thus, our integral will be:
∫[0 to 2π] F(theta) dtheta,
where F(theta) represents the function that defines the area between the curves.
Step 3: Determine the function F(theta).
To determine the function F(theta), we need to calculate the difference between the areas enclosed by each curve at a given value of theta.
The area enclosed by r = (1/2) + sin(theta) is A1 = 1/2 × [θ - sin(theta) cos(theta)].
The area enclosed by r = 0 is A2 = 0 × [θ - sin(theta) cos(theta)], which is zero.
Therefore, the function F(theta) = A1 - A2 = 1/2 × [θ - sin(theta) cos(theta)].
Step 4: Evaluate the integral.
∫[0 to 2π] F(theta) dtheta = ∫[0 to 2π] 1/2 × [θ - sin(theta) cos(theta)] dtheta.
Calculating this integral will give you the area inside the larger loop and outside the smaller loop of the curve r = (1/2) + sin(theta).
Please note that evaluating the integral might involve using numerical methods or software as it can be a challenging task analytically.
To find the area inside the larger loop and outside the smaller loop of the curve specified by r = (1/2) + sin(theta), we can use the concept of double integration in polar coordinates.
First, let's plot the curve to visualize it. We can use a graphing calculator or software like Desmos or Wolfram Alpha for this purpose.
The equation r = (1/2) + sin(theta) represents a polar curve. It consists of two loops: a larger outer loop and a smaller inner loop. The center of the loops is at r = 1/2.
To find the area between the two loops, we need to calculate the area enclosed by the outer loop and then subtract the area enclosed by the inner loop.
To calculate the area between the two loops, we'll integrate the formula for the area of a polar curve over the appropriate range of angles.
Let's set up the integral to find the area of the larger loop:
1. Find the points where the two curves intersect. This can be done by equating the two expressions for r and solving for theta.
(1/2) + sin(theta) = (1/2) - sin(theta)
sin(theta) = -sin(theta)
This is true when either sin(theta) = 0 or -sin(theta) = 0.
For sin(theta) = 0, the solutions are theta = 0 and theta = pi.
For -sin(theta) = 0, the solutions are theta = pi/2 and theta = (3*pi)/2.
So the curves intersect at theta = 0, pi/2, pi, and (3*pi)/2.
2. Set up the integral for the area of the larger loop by integrating the area function over the appropriate range of theta.
To find the area enclosed by the larger loop, we integrate the area function from theta = 0 to pi/2 and then from theta = pi to (3*pi)/2.
Area of larger loop = ∫[(1/2) + sin(theta)]^2 d(theta) from theta = 0 to pi/2
+ ∫[(1/2) + sin(theta)]^2 d(theta) from theta = pi to (3*pi)/2
3. Calculate the integral to find the area between the two loops.
This integral can be challenging to solve analytically, so we can use numerical methods or software to evaluate it. There are many online integral calculators that can provide the result.
After calculating the integral, you should get the desired area inside the larger loop.
To find the area inside the smaller loop, follow the same procedure but integrate the area function only from theta = pi/2 to pi.
Area of smaller loop = ∫[(1/2) + sin(theta)]^2 d(theta) from theta = pi/2 to pi
Finally, subtract the area of the smaller loop from the area of the larger loop to obtain the area between the two loops.
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.