Skip to main content

volume of solid with known cross section

To find the volume of a pyramid with square base we first have to find the dimensions of the pyramid. For this example, the sides of the base are 6 and the height of the pyramid is 4. The lengths of the lines going from the corner of the base square to the top of the pyramid is 5 (found from the distance formula between points (0,4) at the top of the pyramid and (3,0) the corner edge of the rectangle base. This isn't necesarry for the integral but it is helpful in constructing the pyramid in onshape. Next we need to find the equation of the area of a representative rectangle (in this case a square) inside the pyramid. To do this we use the area of a square formula where the sides are "a" length and the area is equal to \(a^2\). We want this in terms of "x" instead of "a", so we say that "a" is equal to "2x". This is because the length from the center of the representative rectangle to the edge of the pyramid is "x" and to account for both sides we get "2x". Thus, in terms of "x", the area of the rectangle is 2\(x^2\). We will find the volume of the pyramid by integrating the area of this rectangle along the height of the pyramid. To do this, we have to get the area equation in terms of y, so we need an equation in terms of y that we can then use to find a relation between x and y and use that relation in terms of x in the area equation. This is slightly confusing but it will make sense once we find an equation in terms of y. To do this we find the equation of the line along the side of the pyramid going from the point at the top of the pyramid to the corner. This gives us \[y=( \frac{-4}{3}x)+4\] because the y-intercept is 4, the rise is -4 and the run is 3.

Now, solve the eqaution for x. \[y=( \frac{-4}{3}x)+4\] \[x= 3-\frac{3}{4}y\] Now we can integrate the equation in terms of y with respect to y \[ \int_{0}^{4} (a^2 )\,dy \] \[ \int_{0}^{4} (2x)^2 \,dy \] \[ \int_{0}^{4} (2(3-\frac{3}{4}y)^2 \,dy \] \[ \int_{0}^{4} (6-\frac{6}{4}y)^2 \,dy \] \[ \int_{0}^{4} 36-18y+\frac{9}{4}y^2 \,dy \] \[volume=48\] So, all we had to do was get the dimensions of the base and height, and we are able to use a slope intercept formula and the formula for the area of a representative rectangle inside the pyramid to find the volume. To aproximate the volume of the pyramid using the rectangles themselves stacked on top of each other and filling the pyramid, we simply add the volume of each rectangle together. The volume of a single rectangle is length*width*height. For this aproximation we will use 10 layers, so the height of each rectangle will be 4in/10layers or 0.4in tall, and the sides of each rectangle will be variable starting with 6in. The sides will decrease in length by 6in/10layers=0.6in/layer each time. We get the below table showing the aproximation of the volume using these rectangles.
To make this representation on Onshape, simply make a a rectangle to represent the base of the pyramid with the dimensions of the base of the pyramid, in this case it is a 1:1 scale so the sides were each 6in. The, to make the first layer extrude it 0.4in because we want 10 layers total and the height of the pyramid is 4in high and 4in/10layers=0.4in/layer. From then on, simply draw another rectangle on top of one another, each time reducing the length by 0.6in, because we want 10 layers and 6in/10layers=0.6in/layer. So, the height for each layer is 0.4in, but the lengths of the rectangles change to be 0.6in less than the one below it. Below are pictures of the resulting representation of these representitive rectangles aproximating the volume of the pyramid on Onshape.
In order to print these in a reasonable dimension, we simply scale the entire pyramid by 1/4 or 0.25 to get a resulting pyramid of height 1 and width 1.5.Shown below.
I chose this shape because I have always been intrigued by pyramids and I think they are the most interesting architectural accomplishment in mankind's design history.
Labels:

Comments

Post a Comment

Popular posts from this blog

Do Over: Integration Over a Region in a Plane

Throughout the semester we have covered a variety of topics and how their mathematical orientation applies to real world scenarios. One topic we discussed, and I would like to revisit, is integration over a region in a plane which involves calculating a double integral. Integrating functions of two variables allows us to calculate the volume under the function in a 3D space. You can see a more in depth description and my previous example in my blog post, https://ukyma391.blogspot.com/2021/09/integration-for-over-regions-in-plane_27.html . I want to revisit this topic because in my previous attempt my volume calculations were incorrect, and my print lacked structural stability. I believed this print and calculation was the topic I could most improve on and wanted to give it another chance. What needed Improvement? The function used previously was f(x) = cos(xy) bounded on [-3,3] x [-1,3]. After solving for the estimated and actual volume, it was difficult to represent in a print...
Post a Comment
Read more

Minimal Surfaces

Minimum surfaces can be described in many equivalent ways. Today, we are going to focus on minimum surfaces by defining it using curvature. A surface is a minimum surface if and only if the mean curvature at every point is zero. This means that every point on the surface is a saddle point with equal and opposite curvature allowing the smallest surface area possible to form. Curvature helps define a minimal surface by looking at the normal vector. For a surface in R 3 , there is a tangent plane at each point. At each point in the surface, there is a normal vector perpendicular to the tangent plane. Then, we can intersect any plane that contains the normal vector with the surface to get a curve. Therefore, the mean curvature of a surface is defined by the following equation. Where theta is an angle from a starting plane that contains the normal vector. For this week’s project, we will be demonstrating minimum surfaces with a frame and soap bubbles! How It Works Minimum surfac...
Post a Comment
Read more

Ruled Surfaces : Trefoil

Ruled Surfaces : Trefoil A ruled surface is a surface that consists straight lines, called rulings, which lie upon the surface. These surfaces are formed of a set of points that are "swept" by a straight line. This is relatively intuitive once you see a good visual, but can be a bit abstract without that concrete example. A very basic example of a ruled surface is a cylinder; if we have a straight line and move it in a circle we create a cylinder made entirely of straight line. Note that the surface will only be a cylinder if all the lines are parallel. If the lines are not parallel we can create hyperboloids and cones depending on how much we have rotated. The rotation we are describing here is not a simple turning action, but more of a twisting motion—less like rotating a can by turning it and more like wringing out a washcloth by twisting it. Specifically, a cylinder is essentially two circles connected by rulings, if we keep one of the circles...
Post a Comment
Read more