The given curve is rotated about the y-axis. Set up, but do not evaluate, an integral for the area of the resulting surface by integrating (a) with respect to x and (b) with respect to y.x = e3y,    0 ≤ y ≤ 2(a)Integrate with respect to x. 1 dx(b)Integrate with respect to y.

The surface area of a solid of revolution can be found using the formula:

A = ∫2πf(x)√[1 + (f'(x))^2] dx

where f(x) is the function being revolved, and f'(x) is its derivative.

(a) Integrating with respect to x:

First, we need to express the function in terms of x. Given x = e^(3y), we can rewrite this as y = (1/3)ln(x). The derivative of y with respect to x is then dy/dx = 1/(3x).

Substituting these into the formula gives:

A = ∫2π[(1/3)ln(x)]√[1 + (1/(3x))^2] dx, from x = e^0 to x = e^6.

(b) Integrating with respect to y:

In this case, the function is already expressed in terms of y, so we can directly substitute into the formula. The derivative of x with respect to y is dx/dy = 3e^(3y).

Substituting these into the formula gives:

A = ∫2π[e^(3y)]√[1 + (3e^(3y))^2] dy, from y = 0 to y = 2.