Evaluate the triple integral ∭ExdV
where E is the solid bounded by the paraboloid x=4y2+4z2 and x=4
.
To solve this triple integral, we will use cylindrical coordinates.
In cylindrical coordinates, the equation of the paraboloid x = 4y^2 + 4z^2 can be written as:
ρ^2 = 4y^2 + 4z^2
Simplifying this equation, we get:
ρ^2 = 4(y^2 + z^2)
This is the equation of a cylinder aligned with the z-axis and centered at the origin, with a radius of √4 = 2.
The equation x = 4 represents a plane that intersects the cylinder at x = 4, forming a circular cross-section.
To evaluate the triple integral, we need to set up the limits of integration.
In cylindrical coordinates, the limits for ρ are from 0 to 2, the limits for θ are from 0 to 2π, and the limits for z are from 0 to the value that satisfies the equation of the plane x = 4:
ρ: 0 ≤ ρ ≤ 2
θ: 0 ≤ θ ≤ 2π
z: 0 ≤ z ≤ (4 - 4y^2)/4
Now, let's calculate the triple integral:
∭ExdV = ∫(0 to 2) ∫(0 to 2π) ∫(0 to (4 - 4y^2)/4) ρ dρ dθ dz
Integrating with respect to ρ, we get:
= ∫(0 to 2) ∫(0 to 2π) [1/2ρ^2] (0 to (4 - 4y^2)/4) dθ dz
= ∫(0 to 2) ∫(0 to 2π) (1/2)(4 - 4y^2)/4 dθ dz
= ∫(0 to 2) ∫(0 to 2π) (1/2)(1 - y^2) dθ dz
Integrating with respect to θ, we get:
= ∫(0 to 2) [θ - (θy^2/2)] (0 to 2π) dz
= ∫(0 to 2) [2π - πy^2] dz
= [2πz - πy^2z] (0 to 2)
= [4π - 2πy^2] - 0
= 4π - 2πy^2
Therefore, the triple integral ∭ExdV = 4π - 2πy^2.
To evaluate the triple integral ∭ExdV, first, we need to express the bounds of integration for each variable.
Given that E is the solid bounded by the paraboloid x=4y^2 + 4z^2 and x=4, we can find the bounds as follows:
1. Bound for x:
Since x=4 is a constant, the lower bound for x is x=0, and the upper bound is x=4.
2. Bounds for y:
The paraboloid equation is x=4y^2 + 4z^2. Rearranging, we get y^2 = (x-4z^2)/4.
Since x=4 is the plane bounding surface E, the equation becomes y^2 = (4-4z^2)/4, or y^2 = 1 - z^2.
Taking the square root of both sides, we have y = ±√(1 - z^2). Thus, the bounds for y are y = √(1 - z^2) and y = -√(1 - z^2).
3. Bounds for z:
Since the paraboloid is defined in the region where x is non-negative, the lower bound for z is z=0. The upper bound can be found by solving the equation x=4y^2 + 4z^2 for z.
Rearranging the equation, we have z^2 = (4-x)/4. Substituting x=4, the equation becomes z^2 = 0, which means z = 0.
Therefore, the bounds for z are z = 0 and z = 0.
Now we can write the triple integral for ∭ExdV as:
∭ExdV = ∫(x=0 to x=4) ∫(y=√(1 - z^2) to y=-√(1 - z^2)) ∫(z=0 to z=0) dx dy dz
Since the upper and lower bounds for z are the same (z=0), the integrand with respect to z simplifies to 1.
∭ExdV = ∫(x=0 to x=4) ∫(y=√(1 - z^2) to y=-√(1 - z^2)) dz dx dy
Integration of the remaining variables can be performed using appropriate limits and the order of integration can be changed based on personal preference or computational convenience.