Evaluate the triple integral ∭ExdV where E is the solid bounded by the paraboloid x=6y^2+6z^2
and x=6
.
To evaluate the triple integral ∭ExdV, we first need to determine the limits of integration for each variable.
The paraboloid equation can be written as x = 6y^2 + 6z^2.
To determine the limits for y and z, we solve for y and z in terms of x:
x = 6y^2 + 6z^2
6y^2 = x - 6z^2
y^2 = (x - 6z^2)/6
y = ±√((x - 6z^2)/6)
Since the paraboloid is symmetric about the yz-plane, we only need to consider the positive square root of y.
Now, we find the limits of integration for y:
When x = 6 (the upper bound), we have:
6 = 6y^2 + 6z^2
y^2 + z^2 = 1
This is the equation of a unit circle in the yz-plane. Therefore, we can let y and z vary over the entire circle, which gives us the limits 0 ≤ y ≤ √1 = 1 and -1 ≤ z ≤ 1.
When x = 6y^2 + 6z^2 = 0 (the lower bound), we have:
0 = 6y^2 + 6z^2
y^2 + z^2 = 0
This equation only holds when y = z = 0. Therefore, the lower bound is a single point (0, 0).
Finally, the limits of integration for x are given by the equation of the paraboloid:
x = 6y^2 + 6z^2
Substituting y = 1 and z = 1, we have:
x = 6(1)^2 + 6(1)^2 = 12
Therefore, the limits of integration for x are 0 ≤ x ≤ 12.
Now we can set up the triple integral:
∭ExdV = ∫(0 to 12)∫(0 to √1)∫(-1 to 1) Exdydzdx
Note that the integrand does not depend on y or z, so we can bring it outside the integral:
∭ExdV = Ex ∫(0 to 12)∫(0 to √1)∫(-1 to 1) dydzdx
Since the integrand does not depend on y or z, the innermost integral becomes:
∫(-1 to 1) dy = 2
The limits of integration for y and z remain the same, so we have:
∭ExdV = 2Ex ∫(0 to 12)∫(0 to √1) dzdx
Now we can evaluate the remaining integrals:
∫(0 to √1) dz = z ∣(0 to √1) = √1 - 0 = √1 = 1
∫(0 to 12) 1 dx = x ∣(0 to 12) = 12 - 0 = 12
Substituting these values back into the expression:
∭ExdV = 2Ex * 1 * 12 = 24Ex
To evaluate the triple integral ∭ExdV, where E is the solid bounded by the paraboloid x = 6y^2 + 6z^2 and x = 6, we need to express the region of integration in terms of y and z.
For the paraboloid x = 6y^2 + 6z^2, we can rearrange to solve for y^2 + z^2:
x - 6z^2 = 6y^2
(x - 6z^2)/6 = y^2 + z^2
This equation represents a solid paraboloid centered at the origin with a vertical axis. The radius of the base is given by sqrt((x-6z^2)/6).
Since the paraboloid intersects the plane x = 6, we have:
6 = 6y^2 + 6z^2
y^2 + z^2 = 1
This equation represents a circle with radius 1 on the xy-plane.
In cylindrical coordinates, we can express the region of integration as:
0 ≤ θ ≤ 2π (for the full circle)
0 ≤ r ≤ √((x-6z^2)/6)
0 ≤ z ≤ √(1 - r^2)
Now, we can set up the triple integral:
∭Ex dV = ∫∫∫E dV
= ∫₀²π ∫₀ᵣ√((x-6z^2)/6) ∫₀²√(1 - r^2) r dz dr dθ
We can now proceed to evaluate this triple integral using appropriate limits and integration techniques.