Evaluate the triple integral ∭ExdV where E is the solid bounded by the paraboloid x=6y2+6z2
and x=6
.
To evaluate the triple integral, we need to determine the limits of integration for each variable: x, y, and z.
The paraboloid is given by x = 6y^2 + 6z^2. We also have the equation x = 6. Setting these two equations equal to each other, we get:
6y^2 + 6z^2 = 6
Dividing both sides by 6, we have:
y^2 + z^2 = 1
This is the equation of a circle with radius 1 centered at the origin in the yz-plane. Therefore, the limits of integration for y and z are -1 to 1.
Since x = 6, the limits of integration for x are from 6y^2 + 6z^2 to 6. Substituting the values for y and z, we have:
6(1)^2 + 6(1)^2 = 12
So the limits of integration for x are from 0 to 12.
Therefore, the triple integral becomes:
∭Ex dV = ∫[from 0 to 12] ∫[from -1 to 1] ∫[from -1 to 1] dxdydz
Evaluating this triple integral, we can integrate with respect to x first:
∫[from 0 to 12] (2)(1)(2) dydz
Simplifying, we have:
∫[from 0 to 12] 4 dydz
Integrating with respect to y next, we have:
4∫[from 0 to 12] dz
Integrating with respect to z, we get:
4z |[from 0 to 12] = 4(12) - 4(0) = 48
Therefore, the value of the triple integral ∭Ex dV is 48.
To evaluate the triple integral of E(x, y, z) dV, we first need to determine the limits of integration for each variable: x, y, and z.
From the given equation x = 6y^2 + 6z^2, we can see that x is constrained between the paraboloid and the plane x = 6. This gives us the following limits for x:
x = 6y^2 + 6z^2 and x = 6
Next, we need to determine the limits for y and z. In this case, we do not have any explicit equations bounding y and z. However, we know that the paraboloid opens upwards, which implies that the surface extends only in the positive y and z directions. Therefore, we will consider y and z to range from 0 to infinity.
Now, we can set up the triple integral:
∭E(x, y, z) dV = ∫[0,∞]∫[0,∞]∫[6y^2 + 6z^2,6] E(x, y, z) dx dy dz
Since E(x, y, z) = 1 (the integrand does not depend on the variables), the integral becomes:
∫[0,∞]∫[0,∞]∫[6y^2 + 6z^2,6] 1 dx dy dz
Now we can integrate with respect to x, y, and z:
∫[0,∞]∫[0,∞]∫[6y^2 + 6z^2,6] 1 dx dy dz
= ∫[0,∞]∫[0,∞] [6 - (6y^2 + 6z^2)] dy dz
Let's start by integrating with respect to y:
∫[0,∞] [6y - 2y^3z^2]∣[0,∞] dz
= ∫[0,∞] (6z) dz
= 6z^2/2∣[0,∞]
Evaluating this integral, we get:
= 3z^2∣[0,∞]
= 3(∞)^2 - 3(0)^2
= 3∞ - 0
= ∞
Therefore, the triple integral ∭E(x, y, z) dV is equal to infinity.