find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis y =(x^2)+1 , y = 9 - (x^2) about y = -1

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find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis y =(x^2)+1 , y = 9 - (x^2) about y = -1

Mathematics
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I like a little visual. |dw:1443749174250:dw|
find the x-coordinates of the intersections there for the limit part
\[\pi[\int\limits_{-2}^{2}(10-x^2)^2-\int\limits_{-2}^{2}(x^2+2)^2\]

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Other answers:

this is what i have so far
i think its wrong because i keep getting the wrong answer
|dw:1443749494904:dw| your radi look right
let me check your x-coordinates of your interesections
\[x^2+1=9-x^2 \\ 2x^2=8 \\ x^2=4 \\ x= \pm 2\]
what did you get as the answer after integration ?
the final answer or do you want me to integrate it?
ill integrate
give the answer and I will check it then if the answer is not what I have I might ask you to show me the integration
i got 864pi for the final answer but wolframalpha says 256pi
ha so it is the integration
\[(10-x^2)^2=100-20x^2+x^4 \\ (2+x^2)^2=4+4x^2+x^4\] you get this for the expansion part?
the difference would be ... \[96-24x^2 \] \[\pi \int\limits_{-2}^{2} (96-24x^2) dx \\ \pi( 96x-8x^3)|_{-2}^{2}\]
\[i get 100x-\frac{ 20x^3 }{ 3 }+\frac{ x^5 }{ 5 }-\frac{ x^5 }{ 5 }-\frac{ 4x^3 }{ 3 }+4x\]
I think that one term should be -4x
not +4x
oh yes i see that
but anyways you should be able to combine like terms after making that correct and get what I have above
it is fine what you did integrate then simplify I prefer to simplify then integrate most of the time
\[[100(2)-\frac{ 20(2)^3 }{ 3 }-\frac{ 4(2)^3 }{ 3 }-4(2)]-[100(-2)-\frac{ 20(-2)^3 }{ 3 }-\frac{ 4(-2) }{ 3 }-4(-2)]\]
ick why not clean up before pluggin in numbers
if you prefer this, this is fine... but it looks gross to me :p
youre right but ive been working on this problem all day so im tryin to be explicit
ok i got it thank you
\[[200-\frac{160}{3}-\frac{32}{3}-8]-[-200+\frac{160}{3}+\frac{32}{3}+8] \\ (200+200)+(-\frac{160}{3}-\frac{160}{3})+(\frac{-32}{3}-\frac{32}{3})+(-8-8) \\ (400)+(\frac{-320}{3})+(\frac{-64}{3})+(-16) \\ \\ (400-16)+(\frac{-320}{3}+\frac{-64}{3}) \\ (384)+(\frac{-384}{3}) \\ 384(1-\frac{1}{3}) \\ 384(\frac{2}{3}) \\ \frac{384}{3}(2) \\ (128)(2) \\ 256\]
my math was just a little off
is the way I would have continued from your last step there
and of course shove the pi in
i got all the integration and the equation and messed up on the algebra
I think if you have simplified before integration would have made it tons easier
il keep that in mind these problems are taking longer and longer
\[\int\limits _{-2}^{2}[(10-x^2)^2-(x^2+2)^2 ]dx\\ \int\limits_{-2}^2 [(100-20x^2+x^4)-(x^4+4x^2+4)] dx \\ \int\limits_{-2}^2 [(100-4)+(-20x^2-4x^2)+(x^4-x^4)] dx \\ \int\limits_{-2}^2 [96-24x^2] dx \\ 96x-24 \frac{x^3}{3} |_{-2}^2 \\ 96x-8x^3|_{-2}^2 \\ (96 \cdot 2 -8(2)^3)-(96 \cdot (-2)-8(-2)^3) \\ (192-64)-(-192+64) \\(192+192)+(-64-64) \\ 384+(-128) \\ 256\]
just in case you wanted to see how I would tackled the integration and the after part

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