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.

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

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.

To find the volume of the solid generated by revolving the region bounded by the curves around the y-axis, we can use the method of cylindrical shells.

The first step is to sketch the region bounded by the curves. In this case, we have the curve y = x^(1/2), y = 0 (the x-axis), and x = 3.

To find the limits of integration, we need to find the points where these curves intersect. Setting y = x^(1/2) equal to 0, we get x = 0 (since the square root of 0 is also 0). Setting y = x^(1/2) equal to x = 3, we can square both sides to get x = 9.

So, the limits of integration for the volume calculation will be from x = 0 to x = 9.

Now, let's consider a vertical strip at a distance x from the y-axis, with width dx. This strip will have an infinitesimal height of Δy.

The circumference of this cylindrical shell is 2πx, and the height of the shell is Δy. Therefore, the volume of this shell is approximately equal to the product of its circumference and height, which is 2πx*Δy.

To find the volume, we integrate this expression over the limits of integration. Since the height of the shell Δy will be given by the difference between the upper and lower curves, we have Δy = x^(1/2) - 0 = x^(1/2).

So, the volume of the solid generated by revolving the region bounded by y = x^(1/2), y = 0, and x = 3 around the y-axis is given by the integral:

V = ∫[0,9] 2πx * (x^(1/2)) dx

Now, we can solve this integral using standard integration techniques.