Find the volume bounded by $C$ and the planes ( integrals ) ( question 9 ) 
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 
 A: 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
$$2\int_{-1}^1\ (1-x)\sqrt{1-x^2}\ dx=\left.\left(\frac{(-2x^2+3x+2)\sqrt{1-x^2}+3\arcsin x}{3}\right)\right|_{-1}^1=\pi.$$
