Find the area between the two spirals r =θ and r = 2θ for 0 ≤ θ ≤ 2Π (two pie). Show your work?

Answer 1

This can be represented as the double integral:
[0, 2π]∫ [θ, 2θ]∫ 1 dA
Making the proper substitution of variables:
[0, 2π]∫ [θ, 2θ]∫ r dr dθ
And integrating:
[0, 2π]∫ (2θ)²/2 - θ²/2 dθ
[0, 2π]∫ 3θ²/2 dθ
(2π)³/2 - (0)³/2
4π³

Steve: Actually, I did subtract in step 4 - note that the minuend is (2θ)²/2, not 2θ²/2. There is an important difference: the two is also being squared.

As for your answer - it cannot possibly be correct. The square [(2π, 0), (2π, -π√2), ((2+√2)π, -π√2), ((2+√2)π, 0)] has area 2π², and is contained entirely within the spirals (indeed, it only occupies a portion of the end), thus the area between them is clearly greater than 2π². Seeing a π³ term might be surprising, but it is quite correct.

Answer 2

In polar coordinates area is

A = ½∫r² dθ

The area between an outer curve r = f(θ) and an inner curve r = g(θ) that do not cross over is

½∫f(θ)² dθ - ½∫g(θ)² dθ.

Here f(θ) = 2θ and g(θ) = θ, so area enclosed is

½∫4θ² dθ - ½∫θ² dθ = ½∫3θ² dθ
= ½ [θ = 0 to 2π: θ³]
= ½ (2π)³
= 4π³

Answer 3

(8 = theta)
This is the intregal of dA = (R-r)d8 where R = 28 & r = 18
Substituting, dA = 8d8, so A = (8^2)over2

Since the integration constant is zero, the area at 8 = 2pi is
(2pi)^2over 2 = 2pi^2

QED

Pascal: Your formula said -, but you added instead of subtracting. Pi cubed? Not too likely for area

Answer 4

two pie over a x plus one forth times two