<s>[INST] Use the disc method to find the volume of solid of revolution. Use the disc method to find the volume of solid of revolution obtained by revolving the the given region around the x-axis.
$R\quad is\quad bounded\quad by\quad the\quad graphs\quad of\quad k(x)\quad =\quad { x }^{ 2 }+x+2\quad and\quad m(x)\quad =\quad 2x+4$
My try :
$ V\quad =\quad \int _{ a }^{ b }{ f(x)^{ 2 } } -g(x)^{ 2 }\quad dx\\ \\ k(x)^{ 2 }\quad =\quad { x }^{ 4 }+2{ x }^{ 3 }+4{ x }^{ 2 }+2x+4\\ m(x)^{ 2 }\quad =\quad 4{ x }^{ 2 }+24x+16\\ \\ \pi \int _{ -1 }^{ 2 } \left| { x }^{ 4 }+2{ x }^{ 3 }-22x-12 \right| \quad dx\\ \\ \left| { \pi \left[ \frac { 1 }{ 4 } { x }^{ 4 }+\frac { 1 }{ 2 } { x }^{ 4 }+11{ x }^{ 2 }-12x \right]  }_{ x=-1 }^{ x=2 } \right| \\ \\ \left| \pi \left[ 4+8-44-24 \right]  \right| -\left| \pi \left[ \frac { 1 }{ 4 } +\frac { 1 }{ 2 } -11+12 \right]  \right| \\ \\ \frac { 217\pi  }{ 4 }  $
The answer in the book is $\frac{369\pi}{10}$
So what did i do wrong ?
--Correction 
$ So\quad i\quad corrected\quad the\quad { k }^{ 2 }\quad and\quad m^{ 2 }\quad \\ \\ k^{ 2 }={ x }^{ 4 }+2{ x }^{ 3 }+5{ x }^{ 2 }+4x+4\\ { m }^{ 2 }=\quad { 4x }^{ 2 }+16x+16\\ \\ \\ \\ \pi \int _{ -1 }^{ 2 }{ \left| { x }^{ 4 }+2{ x }^{ 3 }+{ x }^{ 2 }-12x-12 \right| \quad dx } \\ \\ \left| { \pi \left[ \frac { 1 }{ 5 } { x }^{ 5 }+\frac { 1 }{ 2 } { x }^{ 4 }+\frac { 1 }{ 3 } { x }^{ 3 }-6{ x }^{ 2 }-12x \right]  }_{ x=-1 }^{ x=2 } \right| \\ \\ \left| \pi \left[ \frac { 32 }{ 5 } +8+\frac { 8 }{ 3 } -24-24 \right]  \right| -\left| \pi \left[ \frac { -1 }{ 5 } +\frac { 1 }{ 2 } -\frac { 1 }{ 3 } -6+12 \right]  \right| \\ \\ \frac { 464\pi  }{ 15 } -\frac { 179\pi  }{ 30 } \\ \frac { 749\pi  }{ 30 } $
still wrong answer.
 [/INST] You may want to do these without the absolute value brackets, as I really think that is confusing the issue (and isn't even really correct...).  The line is the "upper function", so the evaluation should be
$$ \pi \ \left[ \ ( - \frac{32}{5} - 8 - \frac{8}{3} + 24 + 24 ) \ - \ (\frac{1}{5} - \frac{1}{2} + \frac{1}{3} + 6 - 12 ) \right] $$
$$ = \ \pi \ (  - \frac{33}{5} + \frac{1}{2} - 8 - 3 + 18 + 36 ) \ = \ \pi \ (43 \ - \ \frac{61}{10} ) \ . $$
You need to be really careful with this evaluation, because the second term is evaluated at negative 1 , so you have to make sure you subtract the odd powers of $ \ x \ $ in the right way.  (I had to re-do this three times myself because there are so many opportunities for sign errors...)
Sign Errors: the #1 cause of math grief, from beginners to professionals!
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