First, we need to find the projection of the cylinder onto the xy-plane, which is just a unit circle centered at the origin. We can then express the integral in cylindrical coordinates as:
∫∫∫ (1-r cosθ) r dz dr dθ
where r is the radius of the circle (i.e. r = 1), θ is the angle from the positive x-axis, and z ranges from the lower bound of z=4+y to the upper bound of z=2+x.
We can express the bounds of z in terms of r and θ by noting that the bottom plane z = 4 + y intersects the cylinder along the curve y = z - 4, so we have:
z = 4 + y = 4 + r sinθ
And similarly, the top plane z = 2 + x intersects the cylinder along the curve x = z - 2, so we have:
z = 2 + x = 2 + r cosθ
We can now set up the integral as follows:
∫∫∫ (1-r cosθ) r dz dr dθ
= ∫₀^²π ∫₀¹ [(1-r cosθ) ∫₍4+r sinθ₎₍²+r cosθ₎ dz] r dr dθ
Evaluating the innermost integral:
∫₍4+r sinθ₎₍²+r cosθ₎ dz
= [(4+r sinθ) - (2+r cosθ)]
= 2 + r sinθ - r cosθ
And plugging that back into the integral:
∫∫∫ (1-r cosθ) (2 + r sinθ - r cosθ) r dz dr dθ
= ∫₀^²π ∫₀¹ (2r + r² sinθ - r³ cosθ - 2r² cosθ + r³ cos²θ) dr dθ
Integrating with respect to r:
= ∫₀^²π [r² + r³ sinθ/3 - r^4/4 - 2/3 r³ cosθ + r⁴/4 cos²θ]₀¹ dr dθ
= ∫₀^²π [1 + sinθ/3 - 1/4 - (8/3) cosθ + 1/4 cos²θ] dθ
Integrating with respect to θ:
= [θ - (8/3) sinθ - (1/3) cosθ + (1/3) sin²θ - (1/12) cos³θ]₀²π
= 4π/3
Therefore, the integral evaluates to 4π/3.
Evaluate the integral of (1-x) over the three dimensional cylinder x^2+y^2=1 bounded below the plane z =4+y and bounded above the plane z=2+x
3 answers
Does theta goes from 0 to 2pi
Yes, you are correct. The limits for θ should be from 0 to 2π instead of 0 to π/2 as I originally wrote. This is because we are integrating over the entire cylinder, which extends through the full range of angles from 0 to 2π. Thank you for pointing out the mistake!
1 answer