Volume of a Pyramid: Find A(z) to Calculate V

[Dustinsfl]

I am trying to find the volume of a pyramid where the base has length \(L\) and width \(W\), and the pyramid has height \(h\).

Let \(L\) be on the x-axis and \(W\) be on the y axis.
In the x-z plane, we have the line \(z = -\frac{h}{L/2}x + h\), and in the y-z plane, we have the line \(z = -\frac{h}{W/2}y + h\).

My cross sections has width \(\Delta z\). So I want to find the volume \(\int_0^hA(z)dz\).

How can I do this?

 

[Evgeny.Makarov]

Hint: Let $A(z)$ be the area of the cross-section at height $z$. Prove that $A(z)=A\left(1-\frac{z}{h}\right)^2$ where $A=A(0)$. Then compute the integral $\int_0^h A(z)\,dz$. This resulting expression of the volume through $A$ works for any right cone, not just a pyramid.

 

[Dustinsfl]

Hint: Let $A(z)$ be the area of the cross-section at height $z$. Prove that $A(z)=A\left(1-\frac{z}{h}\right)^2$ where $A=A(0)$. Then compute the integral $\int_0^h A(z)\,dz$. This resulting expression of the volume through $A$ works for any right cone, not just a pyramid.

So is A a plane? Where did \(\left(1-\frac{z}{h}\right)^2\) come from?

 

[Evgeny.Makarov]

So is A a plane?

No, in my notations $A=A(0)$ is a number, the area of the pyramid's base.

Where did \(\left(1-\frac{z}{h}\right)^2\) come from?

The length of the cross-section at height $z$ is $L\left(1-\frac{z}{h}\right)$, and the width at height $z$ is $W\left(1-\frac{z}{h}\right)$. This is seen from your linear equations by finding $2x$ and $2y$ for a fixed $z$, but it is even easier to see this from similar triangles obtained when you consider sections through the $xz$ and $yz$ planes.

By the way, the formula for the volume in terms of $A$ and $h$ is true for any cone, not necessarily a right one.

 

[Dustinsfl]

No, in my notations $A=A(0)$ is a number, the area of the pyramid's base.

The length of the cross-section at height $z$ is $L\left(1-\frac{z}{h}\right)$, and the width at height $z$ is $W\left(1-\frac{z}{h}\right)$. This is seen from your linear equations by finding $2x$ and $2y$ for a fixed $z$, but it is even easier to see this from similar triangles obtained when you consider sections through the $xz$ and $yz$ planes.

By the way, the formula for the volume in terms of $A$ and $h$ is true for any cone, not necessarily a right one.

Why would we find the area as \(2x\cdot 2y\) instead of \(x\cdot y\)?

 

[Prove It]

I am trying to find the volume of a pyramid where the base has length \(L\) and width \(W\), and the pyramid has height \(h\).

Let \(L\) be on the x-axis and \(W\) be on the y axis.
In the x-z plane, we have the line \(z = -\frac{h}{L/2}x + h\), and in the y-z plane, we have the line \(z = -\frac{h}{W/2}y + h\).

My cross sections has width \(\Delta z\). So I want to find the volume \(\int_0^hA(z)dz\).

How can I do this?

If it's a right pyramid, isn't the volume just $\displaystyle \begin{align*} V = \frac{1}{3}A\,h = \frac{1}{3}L\,W\,h \end{align*}$?

 

[Evgeny.Makarov]

Why would we find the area as \(2x\cdot 2y\) instead of \(x\cdot y\)?

Because the section of the cone using the $xz$ plane is a triangle with sides
\begin{align}
z &= +\frac{h}{L/2}x + h\qquad(1)\\
z &= -\frac{h}{L/2}x + h\qquad(2)
\end{align}
If you find the $x$ corresponding to a given $z$ from (2), then the line at height $z$ crosses the triangle from $-x$ to $x$, i.e., the length of the segment the triangle cuts on the line is $2x$. But again, this is easier to see from similar triangles. The overall intuition is that the length of the cross-section decreases linearly from $L$ at $z=0$ to $0$ at $z=h$. There is a single linear function that does this, and it is $L\left(1-\frac{z}{h}\right)$.

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If it's a right pyramid, isn't the volume just $\displaystyle \begin{align*} V = \frac{1}{3}A\,h = \frac{1}{3}L\,W\,h \end{align*}$?

We are trying to prove it.

 

[MarkFL]

I am trying to find the volume of a pyramid where the base has length \(L\) and width \(W\), and the pyramid has height \(h\).

Let \(L\) be on the x-axis and \(W\) be on the y axis.
In the x-z plane, we have the line \(z = -\frac{h}{L/2}x + h\), and in the y-z plane, we have the line \(z = -\frac{h}{W/2}y + h\).

My cross sections has width \(\Delta z\). So I want to find the volume \(\int_0^hA(z)dz\).

How can I do this?

You may want to read http://mathhelpboards.com/math-notes-49/volumes-pyramids-6131.html for a tutorial on how to work this type of problem. :D

 

[Dustinsfl]

You may want to read http://mathhelpboards.com/math-notes-49/volumes-pyramids-6131.html for a tutorial on how to work this type of problem. :D

I was reading a calculus book and had most of it figured out but the 2x 2y piece Makarov cleared up.