Calculus of a Single Variable, Chapter 7, 7.4, Section 7.4, Problem 46

To find the area of this surface, we rotate the function y = x/2 + 3 about the y-axis (not the x-axis) in the range 1<=x<=5 and this way create a finite surface of revolution.

A way to approach this problem is to swap the roles of x and y , essentially looking at the page side-on, so that we can use the standard formulae that are usually written in terms of x (ie, that usually refer to the x-axis).
The formula for a surface of revolution A is given by (interchanging the roles of x and y )
A = int_a^b (2pi x) sqrt(1+(frac(dx)(dy))^2)dy
Since we are swapping the roles of x and y , we need the function y = x/2 + 3 written as x in terms of y as opposed to y in terms of x . So we have
x = 2y - 6
To obtain the area required by integration, we are effectively adding together tiny rings (of circumference 2pi x at a point y on the y-axis) where each ring takes up length dy on the y-axis. The distance from the circular edge to circular edge of each ring is sqrt(1+(frac(dx)(dy))^2) dy
This is the arc length of the function x = f(y) in a segment of length dy of the y-axis, which can be thought of as the hypotenuse of a tiny triangle with width dy and height dx .
These distances from edge to edge of the tiny rings are then multiplied by the circumference of the surface at that point, 2pi x , to give the surface area of each ring. The tiny sloped rings are added up to give the full sloped surface area of revolution.
We have for this function, x = 2y -6 , that (dx)/(dy) = 2
and since the range (in y ) over which to take the integral is 1 <=x <=5 , or equivalently 7/2 <= y <= 11/2 we have a = 7/2 and b = 11/2 .
Therefore, the area required, A, is given by
A = int_((7)/(2))^((11)/(2)) 2pi (2y -6) sqrt(5) dy = 2sqrt(5)pi int_((7)/(2))^((11)/(2)) 2y - 6 \quad dy

= 2sqrt(5)pi y(y - 6)|_((7)/(2))^((11)/(2)) = (sqrt(5))/(2)pi [11(11-6) - 7(7-6)]
So that the surface area of rotation A is given by
A = 24(sqrt(5)) pi