Question

Question #9747d

Answer 1

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

`=(875pi)/32`

Explanation:

`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

You have not evaluated the integral from 1 to 1.5.

Explanation:

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.