Shells and washers can both be used to measure the volume of a region. Volume is used on a 3-dimensional object. To change a 2-dimensional figure to a 3-dimensional figure we revolve the functions we have around the y-axis or the x-axis. To show this I will be using the shells method with the functions
\[y=x^3-2x\] and \[y=-x^2\] I will be revolving the region between \[0 ≤
x ≤ 1\] about the y-axis. The functions are graphed in the image below.
Using the shells method we would make rectangles parallel to the axis of rotation. For this example we are using the y-axis so the rectangles would be vertical. The more rectangles we make the better the approximation would be for the volume of the object. Revolving these rectangles around the y-axis would make a shell like object. I have made 10 rectangles and revolved it around the y-axis which is shown in the image below. In the image we can see that that the heights of each shell varies and the radius is getting larger by a constant number. The dimensions of the final product will be 4.549 by .909 cubic inches.
I will be measuring the volume of the actual function and I will be comparing it to the volume of the shell model. To measure the actual volume of this function the following formlula would be used \[V = \int_a^b 2*\pi *r*h\] where the r is the radius and h is the height and the points a and b represent the starting and ending points of the region. The radius and the points a and b are all using the x values because we have revolved the region around the y-axis. The height would be the difference of the two functions. The difference can be found by using the function on top and subtracting it with the function on bottom. Using this formula we get \[V = \int_0^1 2*\pi *x*(-x^2-x^3+2) = 1.361\]
Now to compare we can take the volume of each shell to see the difference of the actual volume and our model. The formula to find the volume of the shells is \[V = 2*\pi *r*h*\Delta r\] Where r is the radius, h is the height and \( \displaystyle \Delta\) r is the change of radius (thickness of shell). The table below shows the volume calculated for each shell and the total volume when added together. The total volume approximated for the shells was 1.336. The approximation found from the shells is very close to the actual approximation found which was to be 1.361.
You may ask why were these functions chosen. The first reason is that these functions are easy for someone with a minimal math background to visualize. Being able to visualize the 2-dimensional function can help when trying to visualize the 3-dimensional object. I chose to use two functions because they created steps on the outside of my object and steps on the inside. Being able to see the steps on the outside can be helpful to imagine what is happening on the inside. For someone trying to newly learn this topic these functions would give a good idea on how everything works rather than focusing on the actual math used in the process.
Using the shells method we would make rectangles parallel to the axis of rotation. For this example we are using the y-axis so the rectangles would be vertical. The more rectangles we make the better the approximation would be for the volume of the object. Revolving these rectangles around the y-axis would make a shell like object. I have made 10 rectangles and revolved it around the y-axis which is shown in the image below. In the image we can see that that the heights of each shell varies and the radius is getting larger by a constant number. The dimensions of the final product will be 4.549 by .909 cubic inches.
I will be measuring the volume of the actual function and I will be comparing it to the volume of the shell model. To measure the actual volume of this function the following formlula would be used \[V = \int_a^b 2*\pi *r*h\] where the r is the radius and h is the height and the points a and b represent the starting and ending points of the region. The radius and the points a and b are all using the x values because we have revolved the region around the y-axis. The height would be the difference of the two functions. The difference can be found by using the function on top and subtracting it with the function on bottom. Using this formula we get \[V = \int_0^1 2*\pi *x*(-x^2-x^3+2) = 1.361\]
Now to compare we can take the volume of each shell to see the difference of the actual volume and our model. The formula to find the volume of the shells is \[V = 2*\pi *r*h*\Delta r\] Where r is the radius, h is the height and \( \displaystyle \Delta\) r is the change of radius (thickness of shell). The table below shows the volume calculated for each shell and the total volume when added together. The total volume approximated for the shells was 1.336. The approximation found from the shells is very close to the actual approximation found which was to be 1.361.
You may ask why were these functions chosen. The first reason is that these functions are easy for someone with a minimal math background to visualize. Being able to visualize the 2-dimensional function can help when trying to visualize the 3-dimensional object. I chose to use two functions because they created steps on the outside of my object and steps on the inside. Being able to see the steps on the outside can be helpful to imagine what is happening on the inside. For someone trying to newly learn this topic these functions would give a good idea on how everything works rather than focusing on the actual math used in the process.
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