# Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. y=x^2 , x=y^2; about y=1 Why is the area: A(x)=Pi[(1-x^2)^2-(1-sqrt(x))^2]? Whis is there 1 - X^2 coming from? First the volume of a cylinder is

`pi R^2 h`

If we were to drill a hole through the middle of the cylinder, such that we now had a tube-like thing, or a washer, the volume would be the volume of the original cylinder minus the volume of the hole, or:

`pi R^2 h - pi r^2 h = pi (R^2-r^2) h`

Ok, I don't know how well you can see the various lines in the picture, but here goes:

The black tear drop shape gets rotated around the dotted black line.  If you were to slice this shape with vertical slices, you would get a bunch of washers: cylinders with cylindrical holes in them.  The formula for the volume of this shape is:

`int_0^1 pi (R^2 - r^2) dx`

where R represents the outer radius of a washer slice and r represents the inner radius of a washer slice.

(The `pi(R^2-r^2)`  represents the area of the cross section when you slice it vertically.  The dx can be thought of as the "height" of a washer.  If you are making vertical slices, your cylinders are lying on their sides, and the heights are the `Delta x` .  As you take slices closer and closer together, and let `Delta x -> 0` , this becomes the above integral)

Let's first look at the outer radius:

This is determined by:

The distance from the x axis up to the curve is `x^2`

The distance from the x-axis the the line y=1 is 1

Thus, the distance from the curve to the line y=1 is actually `1-x^2`

So the outer radius is `1-x^2`

(in the first picture, the vertical multicolored line is length 1.  The green piece of it is `x^2` .  Thus the red piece, which is actually the radius, is `1-x^2` )

The distance from the x-axis to the line y=1 is 1 (represented by the multicolored vertical line).

The distance from the x-axis to the curve is `y=sqrt(x)` (represented by the green line)

Thus, the inner radius (represented by the red line) is `1-sqrt(x)`

So, putting that into the formula for volume using washer slices, we have:

`int_0^1 pi[ (1-x^2)^2 - (1-sqrt(x))^2] dx`