Posted by **Danelle** on Saturday, September 1, 2012 at 7:31pm.

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 y-axis.

- math -
**Anu**, Saturday, September 1, 2012 at 7:43pm
Draw graph of y= x^(1/2) to help you understand solution.

Rotation about y-axis means that the radius of the solid (length being swung around the axis is the x-value).

it helps to solve for x-value rather than try to substitute everything. x=y^2 where y>0

each disk or x-value 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 you----its just a variable like x. integrate normally

- math -
**Damon**, Saturday, September 1, 2012 at 7:48pm
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)

- sorry, subtract from total -
**Damon**, Saturday, September 1, 2012 at 7:52pm
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 ]

- math -
**pleh**, Saturday, September 1, 2012 at 7:54pm
thanks

- math -
**Damon**, Saturday, September 1, 2012 at 7:54pm
By the way, you can do it either way.

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