Let $C$ be the cylinder $\big\{(x,y,z):x^2+y^2\leq 1\big\}$. Find the volume bounded by $C$ above the plane $z=0$ and below the plane $x+z=1$.
I have wondered how to solve this , here is my attempt :
$$ \ \int_{x=-1}^{x=1}\int_{z=0}^{z=1-x}\int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}} 1{ \ dy}{\ dz}{\ dx} = \pi$$
i dont know how to solve these kind of questions do i have to sketch the domain ? or to solve it using algebric inequality
Let $Z$ be the finite cylinder $$\big\{(x,y,z)\in\Bbb R^3:x^2+y^2\leq 1\wedge 0\leq z\leq 2\}.$$ Let $B$ denote the required region which is $$\big\{(x,y,z)\in\Bbb{R}^3:x^2+y^2\leq 1\wedge 0\leq z\leq 1-x\big\}.$$ Let $B'$ denote the closure of $Z\setminus B$. Hence, $$B'= \big\{(x,y,z)\in\Bbb{R}^3:x^2+y^2\leq 1\wedge 1-x\leq z\leq 2\}.$$ Then, the following isometry $f:\mathbb{R}^3\to \mathbb{R}^3$ maps $B$ to $B'$: $$f(x,y,z)=(-x,y,2-z).$$ Therefore, $\operatorname{vol}B=\operatorname{vol}B'$. Since $B\cap B'$ is a subset of the plane given by $x+z=1$, $\operatorname{vol}(B\cap B')=0$. Because $Z=B\cup B'$, we obtain $$\operatorname{vol}Z=\operatorname{vol}B+\operatorname{vol}B'-\operatorname{vol}(B\cap B')=2\operatorname{vol}B.$$ As $\operatorname{vol}Z=\pi\cdot 1^2\cdot 2=2\pi$, we get $$\operatorname{vol}B=\pi.$$ This confirms that your solution is correct.
Your solution isn't bad either. Disengaging $y$ and $z$ from your integral, the volume of $B$ is $$2\int_{-1}^1\ (1-x)\sqrt{1-x^2}\ dx=\left.\left(\frac{(-2x^2+3x+2)\sqrt{1-x^2}+3\arcsin x}{3}\right)\right|_{-1}^1=\pi.$$