math
posted by Danelle on .
Find the volume of the solid generated by revolving the region bounded by y= x^(1/2), y= 0, and x= 3 revolving around the yaxis.

Draw graph of y= x^(1/2) to help you understand solution.
Rotation about yaxis means that the radius of the solid (length being swung around the axis is the xvalue).
it helps to solve for xvalue rather than try to substitute everything. x=y^2 where y>0
each disk or xvalue being swung around has an area of pi x^2 (area of a circle equals pi r^2) or in terms of y= pi (y^2)^2= pi Y^4
to make this a volume you need to multiply by the change in y (notation=dy)
so.....integrate
from the interval x=0 to x=3 you need to integrate...when x=0, y==0
when x=3, y= sqrt(3)
so integrate
/sqrt(3) pi(y^4) dy
0/
dont let y confuse youits just a variable like x. integrate normally 
y = x^.5
little cylinder shells of height y and thickness dx
V = int [ 2 pi x y dx } from x = 0 to x = 3
V = int [ 2 pi x^1.5 dx } from x = 0 to x = 3
V = (2 pi/2.5) x^2.5 at 3  at 0
V = (2 pi/2.5)( 3^2.5) 
I did under the curve.
V = pi (3)^2.5  (2 pi/2.5)( 3^2.5)
= 3^2.5 * pi [ 1  2/2.5 ] 
thanks

By the way, you can do it either way.