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 the line `x= -1` .

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Refer to the graph of `y=x^2` and `x=y^2` to determine the region bounded by these two equations.

As shown below, the graph of `y=x^2` is the green curve. And the blue curve is the graph of `x=y^2` . The bounded region is the portion of the graph that is enclosed by the two curves located at the upper right of the graph.

Base on the graph, the two curves intersect at (0,0) and (1,1). Or we may solve this using substitution method of system of equations.

So, substitute `y=x^2` to `x=y^2` .

`x=(x^2)^2`

`x=x^4`

`0=x^4-x`

Factor right side.

`0=x(x^3-1)`

`0=x(x-1)(x^2+x+1)`

Set each factor to zero and solve for x.

`x=0` ,        `x-1=0`                   `x^2+x+1=0`

`x=1`                                `x=(-1+sqrt3i)/2 , (-1-sqrt3i)/2`

Since third factor result to non-real numbers , then we are not going to consider these values of x.

Then, substitute the two values of x, which are 0  and 1,to `y=x^2` .

`x=0` ,           `y=0^2=0`

`x=1` ,           `y=1^2=1`

Hence, the two curves intersect at (0,0) and (1,1).

To determine the volume, plot the axis of rotation x = -1 and the bounded region.

Apply the disk method of volume of solids of revolution. The formula is:

`V = pi int_a^b (r_o^2 - r_i^2)dy`

The limits of the integral are the y-coordinates of the intersection points.

So, a=0 and b=1.

Base on the graph, the inner radius is the distance between the axis of rotation and the blue curve. So,

`r_i = x_(blue)-x_(ax is) = y^2 - (-1)=y^2+1`

Take the square of `r_i` .

`r_i^2 = (y^2+1)^2 = y^4 + 2y^2+1`

And the outer radius is the distance between the axis of rotation and the red curve` y=x^2` .

`r_o= x_(red)-x_(ax is) = sqrty - (-1) = sqrty+1`

Take the square of `r_o` .

`r_o^2= (sqrty+1)^2= y+2sqrty+1 = y+2y^(1/2)+1`

Substitute the expression of `r_i^2` and `r_o^2` to the formula of volume.

`V =pi int_0^1 [(y+2y^(1/2)+1) - (y^4+2y^2+1)]dy`

`V= pi int_0^1(y+2y^(1/2)-y^4-2y^2)dy`

`V=pi (y^2/2 +(4y^(3/2))/3 -y^5/5 - (2y^3)/3)`  `|_0^1`

`V=pi[(1/2 +4/3 - 1/5 - 2/3)-0] = 29/30pi=30.4`

Hence, the volume of the solid obtained by rotating the given equations about x=-1 is 3.04 cubic units.