<s>[INST] 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
 [/INST] I think shells will be easier than washer on this one.
Sketch your region.  We have a line intersecting a parabola, but the parabola is sideways to the normal orientation.
shells $V = 2\pi \int_0^3 y((5-(y-1)^2)-(4-y))\ dy$
$2\pi \int_0^3 3y^2-y^3\ dy\\
2\pi (y^3 - \frac 14 y^4|_0^3) = 2\pi(27 -\frac {81}{4}) = \frac {27}{2}\pi$
by washers.
If you graph your region, you will see a bulge to the right of the points of intersection (4,0) that you must account for.
$V = \pi \int_1^4 (\sqrt{5-x} + 1)^2 - (4-x)^2 \ dx+\int_4^5 (\sqrt{5-x} + 1)^2 - (-\sqrt{5-x} + 1)^2 \ dx$
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