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 $V=\{(x,y,z):{x}^{2}+{y}^{2}\le 4\text{and}0\le z\le 4\}$ 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 ⟨a, b, c⟩ 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?

I have a homework question that goes as follows.

Let $V=\{(x,y,z):{x}^{2}+{y}^{2}\le 4\text{and}0\le z\le 4\}$ 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 ⟨a, b, c⟩ 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?

wlanauee

Beginner2022-10-13Added 17 answers

Step 1

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:

${\iint}_{S}zdxdy$

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.

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:

${\iint}_{S}zdxdy$

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

Beginner2022-10-14Added 4 answers

Step 1

If we let $\overrightarrow{a}=\u27e8-4,4,0\u27e9$ and $\overrightarrow{b}=\u27e8-8,-4,2\u27e9$, then $\overrightarrow{a}\times \overrightarrow{b}=\u27e88,8,48\u27e9$.

Then we can take $\overrightarrow{n}=\u27e81,1,6\u27e9$ 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 $z=\frac{1}{6}(16-x-y)$

If we let $\overrightarrow{a}=\u27e8-4,4,0\u27e9$ and $\overrightarrow{b}=\u27e8-8,-4,2\u27e9$, then $\overrightarrow{a}\times \overrightarrow{b}=\u27e88,8,48\u27e9$.

Then we can take $\overrightarrow{n}=\u27e81,1,6\u27e9$ 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 $z=\frac{1}{6}(16-x-y)$

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