# y = x/2 + 3 , 1<=x<=5 Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the y-axis.

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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

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