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.
im having trouble solving this can someone show me how to do this
The two graphs can be shown as in the figure.
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
outer radius of solid = length from y-axis to y=x/2 grapgh
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`
So the volume of solid encountered by the bound area of `y^2=x` and `2y=x` graphs is 13.404.