Find the volume of revolution when the area between $\displaystyle y=x^2 + 1$ and $\displaystyle y=x+3$ is rotated about the x-axis.

$\displaystyle (x+3)^2 = x^2 + 6x + 9$

$\displaystyle (x^2+1)^2 = x^4 +2x^2 + 1$

So subtract those and I got:

$\displaystyle -x^4 - x^2 + 6x + 8$

Using S for integral as I don't know the code for it.

$\displaystyle V = (pie)S(-1 to 5) (-x^4 - x^2 + 6x + 8)$

Going to jump into the substitution with the anti-deriv.

$\displaystyle =(pie)(-(2)^5 / 5 - (2)^3 / 3 + 3(2)^2 + 8(2)) - (-(-1)^5 / 5 - (-1)^3 / 3 + 3(-1)^2 + 8(-1))$

Got a final answer of $\displaystyle 119/5$ but my friend got a different answer of $\displaystyle 77/5$. What did I do wrong if mine is incorrect