Find the area of one petal of the polar curve r = 3 cos 3θ. (4 points)
A) 7 times pi over 12
B) 3 times pi over 4
C) 11 times pi over 12
D) 2π
Can you please show the workings?
Well, finding the area of a polar curve can be quite the petal-stacking challenge! But fear not, for I am here to help.
To find the area of one petal of the polar curve r = 3 cos 3θ, we can use the formula for the area of a polar region: A = (1/2) ∫[a,b] r^2 dθ.
In this case, the integral becomes A = (1/2) ∫[0, π/6] (3 cos 3θ)^2 dθ.
Simplifying this, we get A = (1/2) ∫[0, π/6] 9 cos^2 3θ dθ.
Using the double angle formula, cos 2θ = 1/2 (1 + cos 4θ), we can rewrite this as A = (1/2) ∫[0, π/6] 9 (1/2) (1 + cos 6θ) dθ.
Expanding and integrating, we have A = (9/4) ∫[0, π/6] (1 + cos 6θ) dθ.
Integrating further, A = (9/4) [(θ + (1/6) sin 6θ) |[0, π/6].
Plugging in the limits, we get A = (9/4) [(π/6 + (1/6) sin (π/3)) - 0].
Simplifying, A = (9/4) [(π/6 + (1/6) (sqrt(3)/2)].
Further simplifying, A = (9/4) [(π/6 + sqrt(3)/12)].
Now let's do some clown math! Simplifying even further, A = (3/2) [(π/6 + sqrt(3)/12)].
Now, let's simplify it down to the finest clown fraction. We have A = (3π + sqrt(3))/8.
So, the area of one petal of the polar curve r = 3 cos 3θ is (3π + sqrt(3))/8.
But hey, let's make it easier to read! We can rewrite it as A = (π + (sqrt(3))/3)/8.
So, the correct answer is E) (π + (sqrt(3))/3)/8, because we all deserve a little slice of clown humor in our math problems!
To find the area of one petal of the polar curve r = 3 cos 3θ, we can use the formula for finding the area enclosed by a polar curve:
A = (1/2) ∫[a,b] (r^2 dθ)
In this case, our limits of integration will be one complete petal, so we can set our limits as θ1 = 0 and θ2 = π/6.
The integral becomes:
A = (1/2) ∫[0,π/6] ((3cos3θ)^2 dθ)
Expanding and simplifying the integral:
A = (1/2) ∫[0,π/6] (9cos^2(3θ) dθ)
Using the identity cos(2θ) = (1/2)(1 + cos(4θ)), we can rewrite our integral:
A = (1/2) ∫[0,π/6] (9/2)(1 + cos(6θ)) dθ
Now we can integrate term by term:
A = (9/4) ∫[0,π/6] (1 + cos(6θ)) dθ
Integrating, we get:
A = (9/4) [θ + (1/6)sin(6θ)] |[0,π/6]
Evaluating the integral at the limits:
A = (9/4) [(π/6) + (1/6)sin(π/2)] - [(0) + (1/6)sin(0)]
Simplifying:
A = (9/4) (π/6)
Simplifying further, we get:
A = (3/4)π/2
Which can be simplified to:
A = 3π/8
Therefore, the area of one petal of the polar curve r = 3 cos 3θ is 3π/8.
Thus, the correct answer is not given among the options.
To find the area of one petal of the polar curve, we can use the formula for finding the area bounded by a polar curve.
The formula for finding the area bounded by a polar curve is as follows:
A = (1/2) ∫[θ1,θ2] (r^2) dθ
In this case, the polar curve is given by r = 3 cos(3θ).
Let's find the values of θ1 and θ2 for one petal of the curve. A petal is formed when the curve completes one full rotation, so we need to find the values of θ for which the curve completes one full rotation.
Since the period of cosine function is 2π, we can set up the equation 3θ = 2π.
Solving for θ, we divide both sides of the equation by 3: θ = (2π)/3.
Therefore, for one petal, θ1 = 0 and θ2 = (2π)/3.
Now, let's substitute the given values into the formula for finding the area:
A = (1/2) ∫[0,(2π)/3] (3 cos(3θ))^2 dθ
Simplifying the expression inside the integral:
A = (1/2) ∫[0,(2π)/3] 9 cos^2(3θ) dθ
We can use the double-angle formula for cosine to simplify further:
cos^2(3θ) = (1/2)[1 + cos(6θ)]
Substituting this into the integral:
A = (1/2) ∫[0,(2π)/3] 9[(1/2)(1 + cos(6θ))] dθ
Now we can simplify the equation inside the integral:
A = (9/4) ∫[0,(2π)/3] (1 + cos(6θ)) dθ
Integrating both terms separately:
A = (9/4)[θ + (1/6)sin(6θ)] from 0 to (2π)/3
Now let's evaluate the integral at the limits:
A = (9/4)[((2π)/3) + (1/6)sin(6((2π)/3))] - [(9/4)(0 + (1/6)sin(6(0))]
Simplifying the expression:
A = (9/4)[((2π)/3) + (1/6)sin(4π)] - 0
Since sin(4π) = 0, the second term simplifies to zero:
A = (9/4)((2π)/3) = 3π/4.
So the area of one petal of the polar curve r = 3 cos 3θ is 3π/4.
Therefore, the correct answer is B) 3 times π over 4.
one petal has -π/6 ≤ θ ≤ π/6
Using the symmetry of the curve, the area is
A = 2∫[0,π/6] 1/2 r^2 dθ = ∫[0,π/6] 9 cos^2(3θ)
= 9/2 ∫[0,π/6] 1+cos(6θ) dθ
Now finish it off