Volume of cylinder cut by two planes

Question

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?

Answer 1

$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 $x$)

$V = 20[0+\frac{9}{2}\frac{\pi}{2}] +0$

$$V = 45\pi $$

Answer 2

The required volume is $$\int_{x=-3}^{3} \int_{y=-\sqrt {9-x^2}}^{\sqrt {9-x^2}} \int _{z=x+y-4}^{1}dz\ dy\ dx.$$

Answer 3

The two planes intersect in the line $x+y=5$ at height $z=1$. This line is outside of the cylinder $x^2+y^2\leq9$. The plane $z=x+y-4$ intersects the axis of the cylinder at $z=-4$, the other plane at $z=1$. The volume of your body therefore is equal to the volume of the standard cylinder of radius $3$ and height $5$, i.e., $V=45\pi$.