Use a triple integral to find the volume of the solid bounded by the parabolic cylinder y=3x^2 and the planes z=0,z=2 and y=1.

To find the volume of the solid bounded by the given surfaces, we can use a triple integral.

The limits of integration for each variable will be determined by the intersection points of the surfaces. Let's break down the problem step by step.

First, let's consider the intersection points of the parabolic cylinder y=3x^2 and the plane y=1. To find these points, set y=3x^2 equal to y=1 and solve for x.

3x^2 = 1
x^2 = 1/3
x = ±√(1/3)

So, the intersection points are (√(1/3), 1) and (-√(1/3), 1).

Next, we need to find the limits of integration for x and z.

The limits of integration for x will be from -√(1/3) to √(1/3) because those are the x-coordinates of the intersection points.

The limits of integration for z will be from 0 to 2 because those are the given planes z=0 and z=2.

Finally, the limit of integration for y will be from y=3x^2 to y=1. Note that the parabolic cylinder y=3x^2 is the lower boundary, and the plane y=1 is the upper boundary.

Now, let's set up the triple integral:

V = ∫∫∫ 1 dy dz dx

The limits of integration are:

x: -√(1/3) to √(1/3)
y: 3x^2 to 1
z: 0 to 2

So, the integral becomes:

V = ∫ from -√(1/3) to √(1/3) ∫ from 3x^2 to 1 ∫ from 0 to 2 1 dy dz dx

Evaluating this integral will give you the volume of the solid bounded by the parabolic cylinder y=3x^2 and the planes z=0, z=2, and y=1.