Angel Kline

2022-10-12

Finding the volume in a cylinder that is intersected by a plane
I have a homework question that goes as follows.
Let 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 $a\left(x-{x}_{0}\right)+b\left(y-{y}_{0}\right)+c\left(z-{z}_{0}\right)=0$ where $\left({x}_{0},{y}_{0},{z}_{0}\right)$ is a point in the plane and ⟨a, b, c⟩ is perpendicular.
I then got two vectors that go between two points, namely:
$PQ=<\left(0-4,\left(4-0\right),\left(2,-2\right)>=<-4,4,0>$ and $PR=<\left(-4-4\right),\left(-4-0\right),\left(4,-2\right)>=<-8,-4,2>$
I then get the cross product of them to get $<8,8,48>$ which is the coefficient in my equation of the plane:
$8\left(x-4\right)+8\left(y-0\right)+48\left(z-2\right)=0$
I think that this is the equation of my plane. Is it?

wlanauee

Step 1
First of all, you made a mistake in your second vector when computing the direction of the plane:
$PR=\left(-8,-4,2\right)$
and your cross product doesn't look correct even with your vector PR.
Now the volume can be set up like this:
${\iint }_{S}zdxdy$
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.
Step 2
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.

Ignacio Riggs

Step 1
If we let $\stackrel{\to }{a}=⟨-4,4,0⟩$ and $\stackrel{\to }{b}=⟨-8,-4,2⟩$, then $\stackrel{\to }{a}×\stackrel{\to }{b}=⟨8,8,48⟩$.
Then we can take $\stackrel{\to }{n}=⟨1,1,6⟩$ as a normal vector to the plane, so substituting in the coordinates of one of the points gives $x+y+6z=16$.
Step 2
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 $z=\frac{1}{6}\left(16-x-y\right)$

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