When calculating the area under a curve, it's easy to split the curve up into slices. These slices are roughly rectangular, and the approximate area can be found by the summation of length times height for each rectangle. If there are two curves, the area between them is simply the area under the top curve minus the area under the bottom curve. This works for two dimensions in the x- and y-planes, but it gets more complicated if these curves get wrapped around an axis and into three dimensions (the z-plane). It is best to visualize the area between two curves in three dimensions as washers or shells.
Take, for example, the curves \[y=25-(x-5)^2 \] and \[ y=(x-5)^2-25 \] wrapped around the y-axis. As seen in the picture, this is a good function to teach with because each of the ten layers on the 3D model coorespond to a whole number on the graph. It's easier for beginners to visualize whole numbers rather than fractions.
The area of this solid of revolution can be found by taking the following integral:
\[ 2pi\int_{0}^{10} (25-(x-5)^2)^2-((x-5)^2-25)^2 dx=(1000/3)*2pi=2094.39 \]
Here, the bounds are where the two curves intersect at 0 and 10. The integral is still the top curve minus the bottom curve; the only difference is that it is multiplied by 2pi to account for the revolution. As seen in the picture, this example was chosen because of its symmetry across the x-axis. This means that another way to calculate the area enclosed by these two parabolas is to take double the area under the first curve bounded by the x-axis as follows:
\[ 4pi\int_{0}^{10} (25-(x-5)^2)^2 dx=(1000/3)*2pi=2094.39 \]
In three dimensions, the rectanglar slices will appear as shells. At x = 1, 2, ... , 10, each shell is a hollowed-out cyclinder. The distinction between viewing these rectangles as shells wrapped around the y-axis instead of washers is that each shell has a constant width. (Washers, on the other hand, would have constant height.) This is depicted in the printed model. Again, because of the symmetry in this example, the model is split into top and bottom halves. This split depicts the x-axis, and we can take the part above and compare it to the part below to see that they are the same. In this model, each shell has a width of 0.2 cm. On the pictured graph, each x- and y-value has been scaled down by 1/5 for the model to give a radius of 1.8 cm and a height of 5 cm for each half. Note that the tenth shell is not visible because it has height zero using right-hand endpoints.
If approximating the area of a solid of revolution using the hollowed-out cylinders called "shells", we can take the equation for the volume of that cylinder and subtract from each shell the volume of a cylinder of the same height with radius r-1. This calculation is pictured for the first two shells. For this example, right-hand endpoints are used.
By taking the sums of the volumes of each shell, the approximated area is 2780.36 units squared. This is close(ish) to the actual value of 2094.39 units squared.
Take, for example, the curves \[y=25-(x-5)^2 \] and \[ y=(x-5)^2-25 \] wrapped around the y-axis. As seen in the picture, this is a good function to teach with because each of the ten layers on the 3D model coorespond to a whole number on the graph. It's easier for beginners to visualize whole numbers rather than fractions.
\[ 2pi\int_{0}^{10} (25-(x-5)^2)^2-((x-5)^2-25)^2 dx=(1000/3)*2pi=2094.39 \]
Here, the bounds are where the two curves intersect at 0 and 10. The integral is still the top curve minus the bottom curve; the only difference is that it is multiplied by 2pi to account for the revolution. As seen in the picture, this example was chosen because of its symmetry across the x-axis. This means that another way to calculate the area enclosed by these two parabolas is to take double the area under the first curve bounded by the x-axis as follows:
\[ 4pi\int_{0}^{10} (25-(x-5)^2)^2 dx=(1000/3)*2pi=2094.39 \]
In three dimensions, the rectanglar slices will appear as shells. At x = 1, 2, ... , 10, each shell is a hollowed-out cyclinder. The distinction between viewing these rectangles as shells wrapped around the y-axis instead of washers is that each shell has a constant width. (Washers, on the other hand, would have constant height.) This is depicted in the printed model. Again, because of the symmetry in this example, the model is split into top and bottom halves. This split depicts the x-axis, and we can take the part above and compare it to the part below to see that they are the same. In this model, each shell has a width of 0.2 cm. On the pictured graph, each x- and y-value has been scaled down by 1/5 for the model to give a radius of 1.8 cm and a height of 5 cm for each half. Note that the tenth shell is not visible because it has height zero using right-hand endpoints.
If approximating the area of a solid of revolution using the hollowed-out cylinders called "shells", we can take the equation for the volume of that cylinder and subtract from each shell the volume of a cylinder of the same height with radius r-1. This calculation is pictured for the first two shells. For this example, right-hand endpoints are used.
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