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 `