Answer to Question #332257 in Calculus for Raul

Question

Find the volume of the solid formed by the region bounded by the functions

y = 6 - 2x - x^2, y = x + 6

Answer 1

$y= 6 -2x -x^2, y = x+6$

Find the intersection:

$\begin{cases} y=6-2x-x^2\\ y= x+6 \end{cases} \rightarrow x+6=6-2x-x^2\\ x^2+3x=0\\ x(x+3)=0\\ x =0, x = -3$

So $V =\pi \int_{-3}^0 (6-2x-x^2)^2-(x+6)^2dx =\pi\int_{-3}^0(x^4+4x^3-9x^2-36x)dx = \pi (\cfrac{x^5}{5}+\cfrac{4x^4}{4} -\cfrac{9x^3}{3}-\cfrac{36x^2}{2}) \Big|_{-3}^0 = -\pi(-\cfrac{243}{5} +81 +81-162) = \pi\cfrac{243}{5}$