<s>[INST] Volume of cylinder cut by two planes I try to find volume of cylinder $x^2 + y^2=9$ cut by inclined plane $z = x + y - 4$ and plane $z=1$
I should solve it in Cartesian, so my idea is to rotate the plane $z=x + y - 4$, and to split the cylinder in two symmetrical portions, which allows using only the portion over $x$ axis. Resulting in two new functions: $j: z=x-2\sqrt{2}$ and  $i: y = \sqrt{9-x^2}$.
My guess is that I can calculate the volume of the cylinder minus the little portion between $z = 0$ and $j$
$V = 2\pi(\text {radius})(\text {height}) - \text {little Area}.$
Can you help me solve this?
 [/INST] $V = \int^3_{x =-3}\int^{\sqrt{9-x^2}}_{y = -\sqrt{9-x^2}}\int^{1}_{z = x+y-4}dzdydx=\int^3_{x =-3}\int^{\sqrt{9-x^2}}_{y = -\sqrt{9-x^2}} \bigg( 1 - x - y +4\bigg)dydx $

$V = \int^3_{x =-3}\bigg(5y-yx-\frac{y^2}{2}\bigg)^{\sqrt{9-x^2}}_{-\sqrt{9-x^2}}dx =\int^3_{x =-3} \big[(5-x)(2\sqrt{9-x^2}) - \frac{1}{2}(9-x^2 - 9 +x^2)\big]dx  $
$V = \int^3_{-3}10\sqrt{9-x^2}dx -\int^3_{-3}2x\sqrt{9-x^2}dx $
$V = 2\cdot 10\bigg[\frac{x}{2}\sqrt{x^2-9}+\frac{9}{2}\sin^{-1}(\frac{x}{3})\bigg]^3_0 + 0 $
(As, $2x\sqrt{9-x^2}$ is an odd function of $x$)
$V = 20[0+\frac{9}{2}\frac{\pi}{2}] +0$
$$V = 45\pi $$
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