Question 203895
I presume you are in a Calculus class. This problem is easier if we use the disc (or cylinder) method of finding the volume of revolution. (Look this up so youcan follow along because I will unable to provide sufficiently clear explanations and/or diagrams to make this self-explanatory.<br>
Below is a graph of y = sqrt(x+1):
{{{graph(600, 600, -2, 13, -2, 5, sqrt(x+1))}}}
The disk method uses the volume of an infinitely thin cylinder (disk). (Picture this disk with the x-axis through the center. Now picture a "stack" of these disks, all centered on the x-axis running from x=2 to x = 9.) To find the volume of all these disks, we sum volumes of each disk over an interval.<br>
The radius of each disk will be the distance from the x-axis (the axis of revolution) to the graph of y = sqrt(x+1). The height of the disk will be "dx". The volume of a cylinder is: {{{V = pi*r^2*h}}}. Replacing "r" with sqrt(x+1) and "h" with "dx" we get, for the volume of one disk:
{{{V = pi*(sqrt(x+1))^2*(dx) = pi*(x+1)*(dx)}}}
The volume of all the disks, from x=2 to x=9, will be the following integral:
{{{int(pi*(x+1), dx, 2, 9) = pi*int((x+1), dx, 2, 9) = pi*(x^2/2 +x)}}} evaluated from 2 to 9:
{{{V = pi*(((9)^2/2 + (9)) - ((2)^2/2 - (2)))}}}
{{{V = pi*((81/2 + 9) - (4/2 - 2))}}}
{{{V = pi*(99/2 - 2)}}}
{{{V = pi*(95/2)}}}
{{{V = (95/2)*pi}}}
I hope this was clear.