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.

Answer 1

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

Answer 2

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)

Answer 3

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 ]

Answer 4

thanks

Answer 5

By the way, you can do it either way.