Asked by John

I need help on this singular question relating to finding the volume of a solid that is obtained. Any help is appreciated, please show work so I can learn better for next time. Thanks!
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"Find the volume of the solid obtained by rotating the region bounded by y=x^2+1 and y=9-x^2
Answered by Tim
I too would like to know the answer to this question!
Answered by John
It is about the x axis, sorry for not including that!
Answered by Anonymous
first sketch graph
where do they cross ?
x^2 + 1 = 9 - x^2
2 x^2 = 8 that is handy !
x^2 = 4
x = 2
y = 5 cross at ( 2 , 5 )
so you want around x axis, integrate from x = 0 to x = 2
upper radius = 9-x^2
lower radius = x^2 + 1
area of outer circle = pi (9-x^2)^2
area of inner circle = pi (x^2 + 1)^2
wee want integral of difference times dx from x = 0 to x = 2
pi * int [ (81 - 18 x^2 + x^4) - (x^4 + 2 x^2 + 1) ] dx from 0 to 2
pi * int [ -20 x^2 + 80 ] dx from 0 to 2
pi [ -20/3 x^3 + 80 x] at 2
pi ( -53 1/3 + 160)
pi (106.66)
about 335

Answered by John
When I tried this, I got exactly double what you got. Why is that?
Answered by John
I think your integral needs to be from -2 to 2, not 0 to 2
Answered by Anonymous
sorry, you are right.
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