Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y^2=x and x=2y about the y-axis. Volume =?im having trouble solving this can someone...

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y^2=x and x=2y about the y-axis.

Volume =?

im having trouble solving this can someone show me how to do this

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

The two graphs can be shown as in the figure.

y=x/2

y=sqrt(x)

The intersections can be obtain;

`sqrtx = x/2`

 `x = x^2/4`

 `0 = x^2-4x`

 `0 = x(x-4)`

 

So the interceptions are at x=0 and x= 4

Coresponding y values are y=0 and y = 2

 

When we rotate the bound area around y, we will get a solid which is like a cone. But it has a internal diameter and external diameter.

Internal diameter is obtained by the y^2 = x graph and outer diameter is obtained by y=x/2 graph.  And y axis is symatrical.

 

Since y-axis is symetrical;

Internal radius of solid = length from y-axis to y^2 = x graph

                                 = y^2

outer radius of solid    = length from y-axis to y=x/2 grapgh

                                 = 2y

 

So the area of a small portion of the solid = `pi[(2y)^2-((y^2)^2)]`

 

Since the integration will be around y-axis the interval will be y cordinates of the intersepting points.

 

Volume = `int A(y)dy`

            = `int_0^2 pi[(2y)^2-((y^2)^2)] dy`

            = `pi*[4y^3/3-y^5/5]_0^2`

            = 13.404

 

So the volume of solid encountered by the bound area of `y^2=x` and `2y=x` graphs is 13.404.

 

Sources:

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