Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

x=3y^2, y=1, x=0 about the y-axis.

To find the volume of the solid obtained by rotating the region bounded by the curves x = 3y^2, y = 1, and x = 0 about the y-axis, we can use the method of cylindrical shells.

First, let's sketch a graph of the given curves and the region they bound:

We have x = 3y^2, which is a parabolic curve opening to the right. The y-axis is the vertical axis, and the region is bounded between y = 0 and y = 1.

Now, let's set up the integral to find the volume using cylindrical shells. The volume of a cylindrical shell is given by the formula:

V = ∫[a,b] 2πr * h * dx

where [a, b] is the interval of integration in terms of x, r is the distance from the axis of rotation to the shell (which is x in this case), h is the height of the shell (which is the difference between the upper and lower bounds of y), and dx is the width of the shell.

The interval of integration in terms of x is from x = 0 to x = 3y^2, which means the interval of y is from y = 0 to y = √(x/3).

The radius of the shell (r) is simply x.

The height of the shell (h) is the difference between the upper and lower bounds of y, which is 1 - 0 = 1.

The width of the shell (dx) can be expressed in terms of y as dx = 2√(3y)dy.

Putting it all together, the integral for finding the volume becomes:

V = ∫[0,1] 2πx * 1 * 2√(3y)dy.

Now, we can solve this integral to find the volume of the solid obtained by rotating the region bounded by the given curves about the y-axis.