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
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.
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!
To evaluate the integral of (1-x) over the given three-dimensional region, we need to set up the limits of integration.
First, let's find the intersection curve of the two planes, z = 4 + y and z = 2 + x, with the cylinder x^2 + y^2 = 1.
At the intersection point, we have:
4 + y = 2 + x
y - x = -2 ---(1)
Since the cylinder is defined by x^2 + y^2 = 1, we can parameterize this curve by letting x = cos(t) and y = sin(t), where 0 ≤ t ≤ 2π.
By substituting these values in Equation (1), we get:
sin(t) - cos(t) = -2
Now, we can find the values of t that satisfy this equation. We'll solve it step-by-step:
sin(t) - cos(t) = -2
sqrt(2) * (sin(t) - cos(t)) = -2 * sqrt(2) ---(2)
sin(t - π/4) = - sqrt(2)/2
Since sin(t) = - sqrt(2)/2 at t = 5π/4 and t = 7π/4, we have:
t = 5π/4 and t = 7π/4
Next, we need to determine the bounds for x, y, and z in the region of integration.
For z, we have:
z = 4 + y and z = 2 + x.
Since the planes are bounded by z = 4 + y and z = 2 + x, the lower bound for z is z = 2 + x, and the upper bound is z = 4 + y.
For x and y, we have:
x^2 + y^2 = 1, x = cos(t), and y = sin(t).
Substituting the values of x and y in the equation x^2 + y^2 = 1, we get:
cos^2(t) + sin^2(t) = 1
This equation is satisfied for all t, so the bounds for x and y are -∞ ≤ x ≤ ∞ and -∞ ≤ y ≤ ∞.
Now, we are ready to set up the integral:
∭(1 - x) dV
∫∫∫(1 - x) dx dy dz
Since the bounds for x, y, and z are -∞ to ∞, we can rewrite the integral as follows:
∫∫∫(1 - cos(t)) dx dy dz
The integral is taken over the region defined by:
0 ≤ t ≤ 2π
-∞ ≤ x ≤ ∞
-∞ ≤ y ≤ ∞
2 + x ≤ z ≤ 4 + y
Now we can evaluate the integral using these limits.