Find the volume of the solid formed by revolving the region bounded by the graph of:

1. x^2+y^2=4 about the line x=3
2. y=6-2x-x^2 and y=x+6

Please help. :(

1)
working with the half above the y-axis, we have y = sqrt(4-x^2); by symmetry we just multiply by an additional factor of 2. Using the Shells method,

dV = 2*pi*r*h*dr

r = 3-x,
h = y = sqrt(4-x^2)
dr = dx

V = 2*(2*pi* integral{-2 to 2, (3-x)*sqrt(4-x^2)dx })
=4*pi* [ 6*arcsin(x/2) - 1/6*sqrt(4-x^2)*(2x^2-9x-8) ] (-2 to 2)

=4*pi*[6*(pi/2 +pi/2) - 1/6*(0 - 0) ]
=24*pi^2.

This can be checked, since it is the volume of a torus, and V = pi^2*r^2*R, with r = 2 and R = 3.

2)

y = 6-2x-x^2
y = 6+x

setting these two curves equal,

x+6-6+2x+x^2 = 3x+x^2 = x(3+x) = 0

they intersect at x = 0,-3.

Rotating about the x-axis, we use the Washer method

dV = pi*(R^2-r^2)*dh

dh = dx
R = 6-2x-x^2
r = 6+x

V = pi*integral{-3 to 0 [ (6-2x-x^2)^2-(6+x)^2 ]dx}
=pi*integral{ -3 to 0 [ (x^4+4x^3-8x^2-24x+36) - (x^2+12x+36) ]dx}
=pi*integral{ -3 to 0 [ x^4+4x^3-9x^2-36x]dx}
=pi* [ x^5/5+x^4-3x^3-18x^2] (-3 to 0)
=pi*(243/5-81-81+162)
=pi*243/5
=152.68