- #1
kira137
- 16
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I've been trying this equation for few hours now and I can't seem to get the right answer..
answer is suppose to be 640(pi)/3
Homework Statement
Find the volume of the solid of revolution obtained by rotating the region bounded by the curves y=x^(2)+1 and y = 9-x^2 around the x-axisThe attempt at a solution
``2
Pi S (9-x^2)^2 - (x^2+1)^2 dx
``-2
``2
Pi S (81-2x^2+x^4-x^4-2x^2-1) dx
``-2
Pi[80x - (4x^3)/3] (x=-2 to 2)
= (448(pi)/3) - (-448(pi)/3) = 896(pi)/3
that's what I keep getting..
and if that isn't the correct answer for rotating around x=0, isn't it suppose to be the answer for when rotating around y=1?
thank you in advance
answer is suppose to be 640(pi)/3
Homework Statement
Find the volume of the solid of revolution obtained by rotating the region bounded by the curves y=x^(2)+1 and y = 9-x^2 around the x-axisThe attempt at a solution
``2
Pi S (9-x^2)^2 - (x^2+1)^2 dx
``-2
``2
Pi S (81-2x^2+x^4-x^4-2x^2-1) dx
``-2
Pi[80x - (4x^3)/3] (x=-2 to 2)
= (448(pi)/3) - (-448(pi)/3) = 896(pi)/3
that's what I keep getting..
and if that isn't the correct answer for rotating around x=0, isn't it suppose to be the answer for when rotating around y=1?
thank you in advance