# Question #9747d

Mar 24, 2017

$\pi {\int}_{- 1}^{\frac{3}{2}} \left[{\left(- {y}^{2} + y - \left(- 4\right)\right)}^{2} - {\left({y}^{2} - 3 - \left(- 4\right)\right)}^{2}\right] \mathrm{dy}$

$= \frac{875 \pi}{32}$

#### Explanation:

$\pi {\int}_{- 1}^{\frac{3}{2}} \left[{\left(- {y}^{2} + y - \left(- 4\right)\right)}^{2} - {\left({y}^{2} - 3 - \left(- 4\right)\right)}^{2}\right] \mathrm{dy}$

$= \pi {\int}_{- 1}^{\frac{3}{2}} \left[{\left(- {y}^{2} + y + 4\right)}^{2} - {\left({y}^{2} + 1\right)}^{2}\right] \mathrm{dy}$

$\textcolor{red}{\text{(FOIL both polynomials separately, then subtract)}}$

$= \pi {\int}_{- 1}^{\frac{3}{2}} \left[- 2 {y}^{3} - 9 {y}^{2} + 8 y + 15\right] \mathrm{dy}$

$= \pi {\left[- \frac{1}{2} {y}^{4} - 3 {y}^{3} + 4 {y}^{2} + 15 y\right]}_{- 1}^{\frac{3}{2}}$

$= \pi \left[- \frac{81}{32} - \frac{81}{8} + 9 + \frac{45}{2} + \frac{1}{2} - 3 - 4 + 15\right]$

$= \pi \left(- \frac{405}{32} + 40\right)$

$= \frac{875 \pi}{32}$

Mar 24, 2017

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

${\int}_{1}^{2} \left({x}^{3} - x\right) \mathrm{dx} = {\left.{x}^{4} / 4 - {x}^{2} / 2\right]}_{1}^{2}$

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

$V = \pi {\left[- {y}^{5} / 5 - \frac{2 {y}^{3}}{3} + \cdot \cdot \cdot - y\right]}_{-} {1}^{1.5}$ you need to substitute $1.5$ into every $y$, the substitute $- 1$ for every $y$ and subtract.