Find a definite integral indicating the area of the surface generated by revolving

the curve
y = 3√3x ; 0 ≤ y ≤ 4 about the x – axis. But do not evaluate the integral.

To find the definite integral indicating the area of the surface generated by revolving the curve y = 3√(3x), 0 ≤ y ≤ 4 about the x-axis, we can use the method of cylindrical shells.

First, we need to express the curve in terms of x, rather than y. Therefore, solving the equation y = 3√(3x) for x, we have:
x = (y/3√3)^2

We are given that 0 ≤ y ≤ 4, so the limits of integration for our integral will be from y = 0 to y = 4.

Now, to find the height of each cylindrical shell, we can recall that the distance from the x-axis to the curve is given by the function itself, in this case, y = 3√(3x).

Next, we need to find the circumference of each cylindrical shell, which can be obtained by considering that when a curve is revolved around the x-axis, the circumference is given by the formula 2πx, where x is the distance from the x-axis to the curve.

Therefore, the circumference of each cylindrical shell is:
C = 2πx = 2π(y/3√3)^2 = (2π/9√3) y^2

Finally, the integral that represents the area of the surface is given by:
A = ∫[0,4] 2πx * h dx, where h is the height of each cylindrical shell and dx is an infinitesimally small change in x.

Substituting the expressions for x and h, we have:
A = ∫[0,4] (2π/9√3) y^2 * (3√(3x)) dx

And that is the definite integral indicating the area of the surface generated by revolving the curve y = 3√(3x), 0 ≤ y ≤ 4 about the x-axis.