help on this one would be appreciated
Find the area of only the area which contains both the circles
r=4cosƒÆ and r=4�ã3sinƒÆ
To find the area of the region bounded by the two given polar curves, we need to determine the points of intersection between the two curves. These points will define the limits of integration for calculating the area.
Step 1: Set the two given polar equations equal to each other:
4cosθ = 4√3sinθ
Step 2: Simplify and solve for θ:
cosθ = √3sinθ
Divide both sides by sinθ:
cotθ = √3
Take the inverse cotangent of both sides:
θ = π/3
Note: The value of θ = π/3 represents the intersection point that lies in the first quadrant. However, since polar coordinates are periodic, we need to find the other two intersections that exist in the third and fifth quadrants.
Step 3: Determine the values of θ for the third and fifth quadrants:
In the third quadrant, θ = π + π/3 = 4π/3.
In the fifth quadrant, θ = 2π - π/3 = 5π/3.
So, we have three intersection points: θ = π/3, 4π/3, and 5π/3.
Step 4: Calculate the area of the bounded region.
The area in polar coordinates is given by the formula:
Area = (1/2)∫[θ1, θ2](r²)dθ
For the bounded region, we need to calculate the area for each separate segment and sum them up.
Segment 1: 0 ≤ θ ≤ π/3
r = 4cosθ
∫[(1/2)(4cosθ)²]dθ
= (1/2)∫[0, π/3](16cos²θ)dθ
Using the double-angle formula cos²θ = (1 + cos(2θ))/2:
= (1/2)∫[0, π/3](16(1 + cos(2θ))/2)dθ
= (1/2)∫[0, π/3](8 + 8cos(2θ))dθ
= (1/2)[8θ + 4sin(2θ)]|[0, π/3]
= (1/2)(8(π/3) + 4sin(2(π/3)) - 0)
= 4π/3 + 2√3
Segment 2: π/3 ≤ θ ≤ 4π/3
r = 4√3sinθ
∫[(1/2)(4√3sinθ)²]dθ
= (1/2)∫[π/3, 4π/3](48sin²θ)dθ
Using the identity sin²θ = (1 - cos(2θ))/2:
= (1/2)∫[π/3, 4π/3](48(1 - cos(2θ))/2)dθ
= (1/2)∫[π/3, 4π/3](24 - 24cos(2θ))dθ
= (1/2)[24θ - 12sin(2θ)]|[π/3, 4π/3]
= (1/2)(24(4π/3) - 12sin(2(4π/3)) - (24(π/3) - 12sin(2(π/3))))
= 24π/3 - 12(-√3) - 8π/3 + 12√3
= 16π/3 + 36√3
Segment 3: 4π/3 ≤ θ ≤ 5π/3
r = 4cosθ
∫[(1/2)(4cosθ)²]dθ
= (1/2)∫[4π/3, 5π/3](16cos²θ)dθ
Using the double-angle formula cos²θ = (1 + cos(2θ))/2:
= (1/2)∫[4π/3, 5π/3](16(1 + cos(2θ))/2)dθ
= (1/2)∫[4π/3, 5π/3](8 + 8cos(2θ))dθ
= (1/2)[8θ + 4sin(2θ)]|[4π/3, 5π/3]
= (1/2)(8(5π/3) + 4sin(2(5π/3)) - (8(4π/3) + 4sin(2(4π/3))))
= 40π/3 - 12√3 - 32π/3 + 12(-√3)
= 8π/3 - 24√3
Now, we can calculate the total area by summing up the areas of all three segments:
Area = (4π/3 + 2√3) + (16π/3 + 36√3) + (8π/3 - 24√3)
Area = (28π/3) + 14√3
So, the area of the region bounded by the two given curves is (28π/3) + 14√3 square units.