Find the volume of the solid obtained by rotating the region bounded by: Q:  "Find the volume of the solid obtained by rotating the region bounded by:
$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$x + y = 4$,  $    $ $    $$x = 5 - (y-1)^2$
$ $ $ $ $ $ $ $ $ $ $ $ $ $ About the $x-axis$."
I have attempted this problem for a while now and keep messing up somewhere. It would much be appreciated if you could point out my mistake and lead me in the right direction.
Attempt: $ $ $ y = 4 - x$, $ $ $y = \sqrt{5-x}+1$,$ $  a= 1 $ $ b= 4
$\pi\int_1^4(\sqrt{5-x}+1)^2-(4-x)^2dx$ =  $\pi\int_1^4-x^2+7x+2\sqrt{5-x}-10 dx$
$\pi(\frac{-x^3}3 +\frac{7x^2}2 +\frac{4/(5-x)^{3/2}}3-10x) |^4_1$  = $\pi(\frac{-64}3+\frac{112}2+\frac{4}3-40+\frac{1}3-\frac{7}2-\frac{32}3+10)$
= -24.60914245312

The equation $x+y=4$ cuts out a cone upon rotation from the solid bounded by the parabola. This occurs for $x\in[1,4]$.For $x>4$, $-\sqrt(5-x)+1>0$ and so also contributes to the integral. You can reduce the work you do by using geometry and subtraction. 
$V=\pi\int_1^5 (\sqrt{5-x}+1)^2  dx$-$\pi\int_4^5 (-\sqrt{5-x}+1)^2 dx - \pi r^2h/3$ where r is the maximum radius of the cone mentioned above, i.e. 3, and h is it's height, also 3.
$$\int_1^5 (5-x)+1+2\sqrt{5-x}=\int_1^5 6-x+2\sqrt{5-x}dx=6x-x^2/2-\frac{4}{3}(5-x)^\frac{3}{2}|_1^5$$
$$=30-\frac{25}{2}-6+\frac{1}{2}+\frac{32}{3}=142\frac{2}{3}$$ 
$$\int_4^5 (1-\sqrt{5-x})^2dx=\int_4^5 6-x-2\sqrt{5-x}dx=6x-x^2/2+\frac{4}{3}\sqrt{5-x}|_4^5$$=$$30-25/2-24+8-\frac{4}{3}=\frac{1}{6}$$
$$V=\pi22\frac{2}{3}-\pi\frac{1}{6}-9\pi=(13\frac{1}{2})\pi$$
