I have a homework question that goes as follows. 

Let $V = \{(x, y, z) : x^2 + y^2 \leq 4 \text{ and } 0 \leq z \leq 4\}$ be a cylinder and let $P$ be the plane through $(4,0,2), (0,4,2)$, and $(-4,-4,4)$. Compute the volume
of $C$ below the plane $P$.
So I have these points and I set up my $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$ where $(x_0, y_0, z_0)$ is a point in the plane and $\langle a,b,c\rangle$ is perpendicular.
I then got two vectors that go between two points, namely: $PQ = <(0-4, (4-0), (2,-2)> = <-4, 4, 0>$ and $PR = <(-4-4), (-4-0), (4,-2)> = <-8,-4,2>$.
I then get the cross product of them to get $<8,8,48>$ which is the coefficient in my equation of the plane:
$$8(x-4) + 8(y-0) + 48(z-2)=0$$ I think that this is the equation of my plane. Is it?
Where do I go from here? I have a feeling I'm supposed to make a triple integral with the function being $8(x-4)  + 8(y-0) + 48(z-2)=0$ and my limits being something to do with the cylinder but I am not sure about it at all. I think my $z$ limits are $0$ to $4$. Is that right?
Thanks a lot.
 A: First of all, you made a mistake in your second vector when computing the direction of the plane:
$$PR=(-8,-4,2)$$
and your cross product doesn't look correct even with your vector $PR$. 
Now the volume can be set up like this:
$$\iint_S z dxdy$$
where $S$ is the projection of the volume to the $xy$ plane and $z$ is the upper limit of $z$ which is the plane represented by $z$.
In fact, you will need to check whether the plane is below $z=4$ in the confined region. After some trick you will see the plane is really below $z=4$. That's why the $z=4$ would not be in your integral any more. 
The triple integral is 
$$\int\int\int_0^z dzdxdy $$
where $z$ is exactly the plane. That is why I wrote the double integral. 
A: If we let $\vec{a}=\langle-4,4,0\rangle$ and $\vec{b}=\langle-8,-4,2\rangle$, then $\vec{a}\times\vec{b}=\langle8,8,48\rangle$.
Then we can take $\vec{n}=\langle1,1,6\rangle$ as a normal vector to the plane, so substituting in the coordinates of one of the points gives $x+y+6z=16$.
Now you can use polar coordinates or rectangular coordinates to find the volume, 
since the base of the solid is the region in the xy-plane bounded by the circle $x^2+y^2=4$, 
and the height of the solid is given by $z=\frac{1}{6}(16-x-y)$.
