Find the area enclosed by r= sin 2θ for 0 ≤θ ≤pi/2. Type your answer in the space below and give 3 decimal places
My answer is: 0.785 is that correct?????
please help me.thanks
almost. In polar coordinates,
A=1/2 ∫ r^2 dθ
check it out here
https://math.libretexts.org/Bookshelves/Calculus/Map%3A_Calculus_-_Early_Transcendentals_(Stewart)/10%3A_Parametric_Equations_And_Polar_Coordinates/10.4%3A_Areas_and_Lengths_in_Polar_Coordinates
Now my answer is different. My answer is 0.3927. Can you check for me please
yes. it is π/8, which is just 1/2 of your previous answer, π/4
To find the area enclosed by the curve r = sin(2θ) for 0 ≤ θ ≤ π/2, we can use the formula for finding the area of a polar region. The formula is given as:
A = (1/2) ∫[a, b] (r(θ))^2 dθ
In this case, we are given the function r(θ) = sin(2θ) and the limits of integration are 0 ≤ θ ≤ π/2. Substituting these values into the formula, we have:
A = (1/2) ∫[0,π/2] (sin(2θ))^2 dθ
Now, we need to evaluate this integral. The integral of (sin(2θ))^2 can be simplified using trigonometric identities:
(sin(2θ))^2 = (1/2) - (1/2)cos(4θ)
Therefore, the integral becomes:
A = (1/2) ∫[0,π/2] [(1/2) - (1/2)cos(4θ)] dθ
Now, we can solve this integral by applying the power rule for integration:
A = (1/2) [(1/2)θ - (1/8)sin(4θ)] evaluated from 0 to π/2
Evaluating this expression at the upper limit and subtracting the evaluation at the lower limit, we get:
A = (1/2) [(1/2)(π/2) - (1/8)sin(2π)] - [(1/2)(0) - (1/8)sin(0)]
Simplifying further, we obtain:
A = (1/2)(π/4) - (1/8)(0) - (0) + (1/8)(0) = π/8
Using a calculator, we can find the numerical value of the area:
A ≈ 0.392 (rounded to 3 decimal places)
Therefore, the correct answer is 0.392, not 0.785.