Use a triple integral to find the volume of the solid enclosed by the paraboloid x=8y^2+8z^2 and the plane x=8.

To find the volume of the solid enclosed by the paraboloid x=8y^2+8z^2 and the plane x=8, we can use a triple integral.

First, let's set up the integral. We need to determine the limits of integration for each variable.

We can solve for y in terms of x and z using the equation of the paraboloid:

x = 8y^2 + 8z^2
x/8 = y^2 + z^2

Since we are given that x=8, we can substitute this into the equation above:
8/8 = y^2 + z^2
1 = y^2 + z^2

This means that the solid is bounded by the sphere with radius 1 centered at the origin.

Next, let's determine the limits of integration for each variable.

For z, since the solid is symmetric about the xy-plane, we can integrate from -1 to 1.

For y, we can integrate from the lower boundary of the radius of the sphere (y = -sqrt(1-z^2)) to the upper boundary of the radius of the sphere (y = sqrt(1-z^2)).

For x, we are given that x=8, so we can integrate from 8 to 8.

Therefore, the triple integral representing the volume of the solid is:

V = ∫∫∫ 1 dx dy dz

Now, we can evaluate this integral:

V = ∫∫∫ 1 dx dy dz
= ∫[8,8] ∫[-sqrt(1-z^2),sqrt(1-z^2)] ∫[-1,1] 1 dx dy dz

Since we have constant limits of integration, we can simplify the integral:

V = ∫[8,8] ∫[-sqrt(1-z^2),sqrt(1-z^2)] 2 dz
= 2 * ∫[8,8] sqrt(1-z^2) dz

To solve this integral, we can make the substitution z = sin(u):

dz = cos(u) du

Now our integral becomes:

V = 2 * ∫[8,8] sqrt(1-sin^2(u)) cos(u) du
= 2 * ∫[8,8] cos(u) cos(u) du

Using the identity cos^2(u) = 1/2 * (1 + cos(2u)), we can simplify the integral further:

V = 2 * ∫[8,8] cos^2(u) du
= 2 * ∫[8,8] 1/2 * (1 + cos(2u)) du
= ∫[8,8] (1 + cos(2u)) du

Now we can just evaluate the integral:

V = ∫[8,8] (1 + cos(2u)) du
= [u + 1/2 * sin(2u)] [8,8]
= 8 - 8 + 1/2 * sin(2*8) - 1/2 * sin(2*8)
= 0

Therefore, the volume of the solid enclosed by the given paraboloid and plane is 0.

To find the volume of the solid enclosed by the paraboloid x=8y^2+8z^2 and the plane x=8, we can use a triple integral with appropriate limits of integration.

First, let's find the limits of integration for each variable by considering the given equations.

Since the plane x=8 intersects the paraboloid, we can set the equation of the paraboloid equal to 8:
8y^2 + 8z^2 = 8

Dividing both sides by 8, we get:
y^2 + z^2 = 1

This equation represents a 2D circle centered at the origin in the yz-plane. We can use this information to determine the limits of integration for y and z.

In cylindrical coordinates, the equation of the circle becomes:
r^2 = 1

So, the limits of integration for r are 0 to 1.

To find the limits of integration for z, we need to determine the height of the paraboloid at each value of y. Let's rewrite the equation of the paraboloid to solve for x:
x - 8y^2 - 8z^2 = 0

Since x = 8, we have:
8 - 8y^2 - 8z^2 = 0

Simplifying, we get:
-8y^2 - 8z^2 = -8

Dividing by -8, we obtain:
y^2 + z^2 = 1

This is the same equation as before, which means the paraboloid intersects the plane x = 8 at the circle y^2 + z^2 = 1. Therefore, the limits of integration for z are -√(1-y^2) to √(1-y^2).

Lastly, we need to determine the limits of integration for y. Since the circle y^2 + z^2 = 1 lies within the range -1 ≤ y ≤ 1, we can conclude that the limits of integration for y are -1 to 1.

Now, we can set up the triple integral to find the volume enclosed by the paraboloid and the plane. The volume V is given by:

V = ∬∬∬ dV

Where dV represents an infinitesimally small volume element.

The limits of integration for y, z, and r are -1 to 1, -√(1-y^2) to √(1-y^2), and 0 to 1, respectively. Thus, the triple integral becomes:

V = ∫[0 to 1] ∫[-1 to 1] ∫[-√(1-y^2) to √(1-y^2)] r dz dy dr

Evaluating this triple integral will give us the volume of the solid enclosed by the paraboloid and the plane x=8.