Question #9747d

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Answer 1 · 2017-03-24T03:16:56

$pi int_(-1)^(3/2)[(-y^2+y-(-4))^2-(y^2-3-(-4))^2]dy$

$=(875pi)/32$

$pi int_(-1)^(3/2)[(-y^2+y-(-4))^2-(y^2-3-(-4))^2]dy$

$=pi int_(-1)^(3/2)[(-y^2+y+4)^2-(y^2+1)^2]dy$

$color(red)("(FOIL both polynomials separately, then subtract)")$

$=pi int_(-1)^(3/2)[-2y^3-9y^2+8y+15]dy$

$=pi [-1/2y^4-3y^3+4y^2+15y]_(-1)^(3/2)$

$=pi [-81/32-81/8+9+45/2+1/2-3-4+15]$

$=pi(-405/32+40)$

$=(875pi)/32$

Answer 2 · 2017-03-24T15:55:33

You have not evaluated the integral from 1 to 1.5.

I don't want to re-type your entire answer, but you have done the equivalent of

$int_1^2 (x^3 - x) dx = {: x^4/4 - x^2/2 ]_1^2$

And the you have gone back to

$x^3+x$ but you've substituted the upper limit for only the first $x$ and the lower for the second.

$2^4/4-1^2/2$ .

Once you get

$V = pi[-y^5/5-(2y^3)/3 + * * * -y]_-1^1.5$ you need to substitute $1.5$ into every $y$ , the substitute $-1$ for every $y$ and subtract.