Finding the volume using double integrals Find the volume  of the wedge sliced from the cylinder $x^2+y^2 =1$ by the planes $z=a(2-x)$ and $z=a(x-2)$  $a>0$
I am confused because $x^2+y^2 = 1$ is a unit circle not a cylinder. and the other two are lines in the $zx$ plane where $y=0$. I don't see how they are planes, are they planes or are they just lines in the $zx$ plane (when $y=0$). I tried drawing the 3d picture but it made no sense since a cylinder wasn't formed.
 A: In a figure showing the $(x,z)$-plane the two given planes appear as lines intersecting the $x$-axis at $x=2$ and the $z$-axis at $z=\pm2a$. Looking at the figure one realizes that because of symmetry the volume in question is $4a\pi$, whereby $\pi$ stands for the area of the circle $x^2+y^2\leq1$..
A: Notice that there are many x-y planes (see graphic).

A: It is clear that the problem is given in $\mathbb{R}^3$. Therefore when someone says "the cylinder $x^2+y^2=1$" they mean to indicate the following set:
$$\{(x,y,z)\in\mathbb{R}^3: x^2+y^2=1\}$$
which is an infinite cylinder obtained by sweeping the circle $x^2+y^2=1$ up and down vertically in the $z$-direction, that is, perpendicular to the plane $z=0$.
Now the general equation of a plane in $\mathbb{R}^3$ is this:
$$Ax+By+Cz = D$$
for some $A,B,C,D\in \mathbb{R}$. Choosing $A=-a,B=0,C=1,D=2a$, gives the plane $z=a(2-x)$. It is a plane perpendicular to the vector $(-a,0,1)$ that lies a distance of $2a$ from the origin. If you take $A=a, B=0, C=1, D=-2a$, you get the other plane, namely $z=a(x-2)$, which is perpendicular to the vector $(a,0,1)$ and also lies at distance of $2a$ from the origin.
So you got the entire geometry mixed up. Try it now.
