Solids of revolution help please

Please set up the integral for the volume of the solid formed by rotating the region bound by y=x and y=sqrt(x) about y=-1

I have no idea what to do with the "y=-1" part :((((((

For this you're going to use the basic "washer" method.

Pi * Integral (a,b) [(radius 1)^2 - (radius 2)^2 ] dx

For this, the "top" radius is going to be the distance between y=sqrt (x) and y=-1
{radius 1} = y = sqrt(x) - (-1) = sqrt(x) + 1

The "bottom" radius is the distance between y=x and y=-1
{radius 2} = x+1

The limits are at x=0 and x=1

So the integral is:

Pi* Integral(0,1) [(sqrt(x)+1)^2 - (x+1)^2] dx
which can be written as
Pi * Integral(0,1) [-x^2-x+2*sqrt(x)] dx
The antiderivative is:
Pi * [-1/3 x^3 - 1/2x + 4/3 x^3/2] | (0,1)

Evaluate to get:
1/2

Remember, the y = -1 part is just going to change what your radii are in the initial equation. Sometimes it's easier to see if you graph the equation

First graph the Region .- -1= sqrt x , x= 1
The radius of revolution id R= (-1 ) - f(x) , then the volume is
dV = piR^2 dx

V1= pi INT (-1-x)^2 dx -1 V2 = pi INT ( -1-sqrtx)^2 dx 0
V=V1+V2
V1= (pi/3) (1+x)^3 = (pi/3) (1^3-0^3)

V2= pi ( INT dx + 2INT sqrtx dx + INT xdx )
V2= pi (x + (4/3)x^(3/2) +x^2/2 )
V2= pi ( 1+(4/3)+1/2 -0)
V2 = 17pi/6
V= pi/3+17pi/6 = 19pi/6