[SOLVED]Volume of a pyramid

dwsmith

Well-known member
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

Well-known member
MHB Math Scholar
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.

dwsmith

Well-known member
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

Well-known member
MHB Math Scholar
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.

dwsmith

Well-known member
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

Well-known member
MHB Math Helper
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

Well-known member
MHB Math Scholar
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}
\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 $L\left(1-\frac{z}{h}\right)$.

- - - Updated - - -

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

Staff member
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 >>>this thread<<< for a tutorial on how to work this type of problem. dwsmith

Well-known member
You may want to read >>>this thread<<< for a tutorial on how to work this type of problem. I was reading a calculus book and had most of it figured out but the 2x 2y piece Makarov cleared up.