Volume Problem - Riemann Sum

In summary: This gives us:s(y)=\frac{s}{h}y+\frac{s}{2h}This equation can be rewritten in slope-intercept form:y-y_0=s_0+h_0s_0s_0=-s_0h_0Hence:s(y)=-\frac{s}{h}y+\frac{s}{2h}
  • #1
MathsKid007
3
0
THE QUESTION
By using Riemann’s sum, synthesise a mathematical model for finding the exact volume of any ‘tepee’ tent of side s and height h.
HERE'S WHAT I HAVE
View attachment 8289
Am currently stuck on writing a side length for the hexagon at any height 'x'
 

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  • #2
Let's first find the volume using the geometric formula:

\(\displaystyle V=\frac{1}{3}bh\)

Now, the base is a regular hexagon having side lengths \(s\), which we can decompose into 6 equilateral triangles:

\(\displaystyle b=6\left(\frac{1}{2}s^2\sin\left(\frac{\pi}{3}\right)\right)=\frac{3\sqrt{3}}{2}s^2\)

Hence:

\(\displaystyle V=\frac{\sqrt{3}}{2}hs^2\)

Now, using volume by slicing we can integrate:

\(\displaystyle V=\frac{3\sqrt{3}}{2}\int_0^h s(y)^2\,dy\)

We know \(s(y)\) will vary linearly, where:

\(\displaystyle s(0)=s\)

\(\displaystyle s(h)=0\)

Thus:

\(\displaystyle s(y)=-\frac{s}{h}y+s\)

And so we may write:

\(\displaystyle V=\frac{3\sqrt{3}}{2}\int_0^h \left(-\frac{s}{h}y+s\right)^2\,dy\)

At this point, I would consider the substitution:

\(\displaystyle u=-\frac{s}{h}y+s\implies du=-\frac{s}{h}dy\)

\(\displaystyle V=\frac{3\sqrt{3}h}{2s}\int_0^s u^2\,du=\frac{3\sqrt{3}h}{2s}\left(\frac{s^3}{3}\right)=\frac{\sqrt{3}}{2}hs^2\quad\checkmark\)

It appears you are directly integrating rather than forming a Riemann sum. Is that on purpose, or are you integrating first, and then going to formulate the sum?
 
  • #3
Could u explain the process at:
We know that s(y) will vary linearly such that:
s(0)=s
s(h)=0
∴s(y)= -s/h y+s
Thanks :D
 
  • #4
MathsKid007 said:
Could u explain the process at:
We know that s(y) will vary linearly such that:
s(0)=s
s(h)=0
∴s(y)= -s/h y+s
Thanks :D

We know that at the bottom, where \(\displaystyle h=0\), the length of the sides of the hexagonal cross section is \(\displaystyle s\). We also know that at the top these sides lengths have diminished to 0. As they have diminished linearly, and we know two points on this line we have all we need to determine the side lengths as a function of the variable height \(\displaystyle y\) using the definition of slope and the point-slope formula:

\(\displaystyle s(y)-s=\frac{0-s}{h-0}(y-0)\)

\(\displaystyle s(y)=-\frac{s}{h}y+s\)
 

What is the volume problem in relation to Riemann Sum?

The volume problem in relation to Riemann Sum is a concept in calculus that involves finding the volume of a 3-dimensional shape by using a series of rectangles with varying widths and heights. It is based on the Riemann Sum, which is a method for approximating the area under a curve.

How is the Riemann Sum used to solve the volume problem?

The Riemann Sum is used to solve the volume problem by dividing the shape into smaller rectangles with known widths and heights. The sum of the volumes of these rectangles can then be used to approximate the volume of the entire shape. As the number of rectangles used increases, the approximation becomes more accurate.

What is the difference between left, right, and midpoint Riemann Sum in the context of the volume problem?

In the context of the volume problem, the left, right, and midpoint Riemann Sum refer to different ways of approximating the volume of a shape using rectangles. Left Riemann Sum uses the left endpoint of each rectangle as its height, right Riemann Sum uses the right endpoint, and midpoint Riemann Sum uses the midpoint between the left and right endpoints. These different methods can result in slightly different approximations of the volume.

What are the limitations of using Riemann Sum to solve the volume problem?

One limitation of using Riemann Sum to solve the volume problem is that it can only approximate the volume of shapes with flat surfaces, such as cubes and cylinders. It cannot be used for shapes with curved or irregular surfaces. Additionally, as the number of rectangles used increases, the calculation becomes more complex and time-consuming.

How is the Riemann Sum used in real-life applications?

The Riemann Sum has many real-life applications, such as calculating the volume of objects in engineering and architecture, determining the area under a curve in physics and economics, and estimating the total distance traveled in a moving object's motion. It is also used in computer graphics to render 3D images and animations.

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