Find the volume of a solid generated by revolving the graphs of the functions y=6-2x-x^2 and y=x+6 about the line y=2
Intersections points are x=-3 and x=0
I think the radius of the solid should be 6-2x-x^2-(x+6)-2 which is -2-3x-x^2
then I have Vol = pi*Integral from -3 to 0 of (-2-3x-x^2)^2dx
I integrate and get pi *[4x65/5+3/2x^4+13x^3/3+6x^2+4x ] evalulated between -3 and 0. I end up with 147.9pi which seems way too big.Can you find my mistake? Thank you so much!!!

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if i see it right; you went ahead and subtracted 2 to get it to the x axis right?

y = 4 -2x -x^2
y = x+4 and spin that around the x axis (y=0)

is your interval good?
6-2x-x^2 -x-6
-x^2 -3x = 0
-x(x+3) = 0; x= [-3,0] is good

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