A cardioid r=1+cos(theta)
A circle r=3*cos(theta)
a) Define the domain of the region enclosed inside both the cardioid and the circle.
b) Use polar coordinates to calculate the area.
(We can use symmetry about x-axis)
May I suggest two excellent videos from the
Khan Academy
The first develops the standard method of finding the area
https://www.khanacademy.org/math/integral-calculus/solid_revolution_topic/area-polar-graphs/v/formula-area-polar-graph
And the second just about matches your question
https://www.khanacademy.org/math/integral-calculus/solid_revolution_topic/area-polar-graphs/v/cardioid-area
The videos did help but i still have trouble setting up the integrals. Since this is part of double integrals section in my course... i'm trying to do it in the form of double integrals.
Is this correct?
Outer integral is from -pi/2 to pi/2
Inner is [r dr d(theta)] from 3*cos(theta) to 1+cos(theta) ?
thanks
a) To find the domain of the region enclosed inside both the cardioid and the circle, we need to determine the range of values for theta that satisfies both equations.
Starting with the cardioid equation r = 1 + cos(theta):
1 + cos(theta) >= 0
cos(theta) >= -1
-1 <= cos(theta) <= 1
Since the cosine function has a range of [-1, 1], this means that the equation is satisfied for all values of theta.
Now, let's consider the circle equation r = 3 * cos(theta):
3 * cos(theta) >= 0
cos(theta) >= 0
Since the cosine function is positive for angles between 0 and π, we can conclude that the equation is satisfied for theta in the range [0, π].
Combining the conditions from both equations, we get the domain of the region enclosed to be:
Domain: 0 <= theta <= π
b) To calculate the area using polar coordinates, we can integrate the equation for the region inside both the cardioid and the circle, with respect to theta.
Since we have symmetry about the x-axis, we can calculate the area for half of the region and then double it.
The area can be calculated using the following integral:
A = 2 * ∫[0, π] (1/2) * ((3 * cos(theta))^2 - (1 + cos(theta))^2) d(theta)
Simplifying the equation:
A = 2 * ∫[0, π] (1/2) * (9 * cos^2(theta) - (1 + 2 * cos(theta) + cos^2(theta))) d(theta)
A = ∫[0, π] (1/2) * (8 * cos^2(theta) - 2 * cos(theta) - 1) d(theta)
Now, we can integrate the equation within the given range to calculate the area.
To define the domain of the region enclosed inside both the cardioid and the circle, we need to determine the range of values for the angle θ.
a) For the cardioid r = 1 + cos(θ), we know that cos(θ) is in the range [-1, 1]. Therefore, the range of values for 1 + cos(θ) is [0, 2]. So, for the cardioid, the domain of θ is [0, 2π], covering a full revolution.
b) For the circle r = 3cos(θ), we know that cos(θ) is in the range [-1, 1]. Therefore, the range of values for 3cos(θ) is [-3, 3]. Since r = 3cos(θ), we have that the radius of the circle can be negative as well. However, we want to find the region enclosed inside the circle, which means we only consider the positive values. So, for the circle, the range of values for θ is [0, π], covering half a revolution.
Now, to calculate the area using polar coordinates, we can divide the region into two areas, one above and one below the x-axis, and use symmetry about the x-axis to calculate the full area.
We can calculate the area using the formula:
Area = ∫[θ1,θ2] (1/2) r^2 dθ
In this case, we will integrate twice, once for each area:
For the area above the x-axis:
Area1 = ∫[0, π] (1/2) (1 + cos(θ))^2 dθ
For the area below the x-axis, we can use the symmetry property:
Area2 = 2 * Area1 = 2 * ∫[0, π] (1/2) (1 + cos(θ))^2 dθ
Now, we can calculate the areas using numerical integration methods or using calculus techniques, such as integration by substitution or integration by parts.