Volume of a random pyramid

In this problem, you’re going to find the volume of your own, personal pyramid. To find the specific problem, choose your name from the following list:

My pyramid has height 2 and a base that’s 5 by 4. When the slice is raised up off of the base a bit, it looks like so:

Screen Shot 2022-02-13 at 12.33.54 PM1168×684 10.6 KB

Let’s express the cross-sectional area function A(z) in terms of a scaling function s(z) that tells a linear scaling factor as a function of height z. Clearly,

since the slice is full sized at the bottom and zero sized at the top. Furthermore, since the sides of the pyramid are straight lines, it would seem that s(z) should have the form

Since we know how to find the equation of a line given two points, it’s not too hard to see that

Now, the area of the base is 5\times4=20 so the cross-sectional area function should be

Thus, the volume is

My pyramid has a base length of five and a width of two, with a height of three.
So s(0)=1, s(3)=0
and because we have straight lines, we can represent those lines with the equation s(z)=mz+c
Therefore
s(z)=-\frac{1}{3}z+1=1-\frac{z}{3}
The base of our pyramid is 2*5=10, so we apply our cross section formula of A(z)=base(slope)^2, which gives us
A(z)=10(1-\frac{z}{3})^2
in order to find volume, we need to integrate that equation over our height.
10\int_0^3(1-\frac{z}{3})^2dz=10

The horizontal cross-sectional area at each slice is A = 5s*4s = 20s^2, as s is the scaling factor as the slice is moving from the bottom to the top.
If we slice diagonally across the z-axis, we see that:
\frac{r}{R}=\frac{h-z}{h}, as r is the diagonal length of the horizontal cross-sectional surface, R is the diagonal length of the base, h is the height of the pyramid, z is the height of the cross-sectional area from the base.
Thus, s=\frac{r}{R}=1-\frac{z}{h}=1-\frac{z}{6}
The volume of the pyramid is V=20\int_0^6(1-\frac{z}{6})^2dz=20\int_0^6(1-\frac{z}{3}+\frac{z^2}{36})=20(z-\frac{z^2}{6}+\frac{z^3}{108})|_0^6\\=20(6-6+\frac{216}{108})=40

My pyramid has a height 5 and a base that’s 3 by 5.
So s(0)=1 and s(5) =0
With straight lines we can represent them with
s(z) = mz+c
Meaning
s(z) = - \frac{1}{5} z+1 = 1 - \frac{z}{5}
With a base area of 5*3 =15 we can apply the cross sectional area formula and get
A(z) = 15(1- \frac{z}{5})^2
Finally, we find the volume to be
15\int_0^5 (1- \frac{z}{5})^2 dz = \frac{5}{3}.

The cross-sectional area function A(z) in terms of scaling function s(z) that tells a linear scaling factor as a function of height z. So
s(0) = 1 \text{ and } s(7) = 0,
Since the slice is full-sized at the bottom and zero-sized at the top. That would seem that s(z) have the form of
s(z)=mz+c
Since we know how to fund the equation of a line given two points, it’s not too hard to see that
s(z) = -1/7z+ 1 = 1 - z/7
Now, the area of the base is 6x4 = 24 so the cross-section area function should be
A(z) = 24(1-z/7)^2
Thus the volume is
24\int_0^7 (1-z/7)^2 dz = 56.