The washer method is to slice an object into washer shaped slices, and
then integrate over these slices to compute the volume of the object.
By rotating an ellipse around the y-axis, we can generate a torus whose volume can be calculated using the washer method.
The function of the selected ellipse is as following: \[\frac{(x-0.6)^2}{0.2^2}+\frac{(y-0.5)^2}{0.5^2}=1\] We can get a torus generated by rotating the selected ellipse around the y-axis. It is feasible to use washer shaped slices stacked along the y-axis to approximate the torus. We can calculate the area of the ring for each washer shaped slice, and then we compute the volume of the washer shaped slice by multiplying the area and the height of the washer shaped slice. The calculation process is shown in the following table: From the table above we got the approximation of the volume is 1.145 cubic inches.
Use the following integral to calculate the real volume of the model \[V=\pi \times \int_0^1{\{(0.6+0.2 \times \sqrt{(1-4(y-0.5)^2 })^2-(0.6-0.2\times \sqrt {(1-4(y-0.5)^2 })^2 }\}dy\\ =\pi \times \frac{3(arcsin(2y-1)+\sqrt{(1-y)y}(4y-2))}{25}\bigg|_0^1\\ =\pi \times \frac{3\pi}{25}\\ =1.184\] The real volume of the torus is 1.184 cubic inchs. The difference between the approximated volume and the accurate volume is 1.184-1.145=0.039 cubic inches. The relative error is 0.039/1.184 ≈ 3.3%.
By rotating an ellipse around the y-axis, we can generate a torus whose volume can be calculated using the washer method.
The function of the selected ellipse is as following: \[\frac{(x-0.6)^2}{0.2^2}+\frac{(y-0.5)^2}{0.5^2}=1\] We can get a torus generated by rotating the selected ellipse around the y-axis. It is feasible to use washer shaped slices stacked along the y-axis to approximate the torus. We can calculate the area of the ring for each washer shaped slice, and then we compute the volume of the washer shaped slice by multiplying the area and the height of the washer shaped slice. The calculation process is shown in the following table: From the table above we got the approximation of the volume is 1.145 cubic inches.
Use the following integral to calculate the real volume of the model \[V=\pi \times \int_0^1{\{(0.6+0.2 \times \sqrt{(1-4(y-0.5)^2 })^2-(0.6-0.2\times \sqrt {(1-4(y-0.5)^2 })^2 }\}dy\\ =\pi \times \frac{3(arcsin(2y-1)+\sqrt{(1-y)y}(4y-2))}{25}\bigg|_0^1\\ =\pi \times \frac{3\pi}{25}\\ =1.184\] The real volume of the torus is 1.184 cubic inchs. The difference between the approximated volume and the accurate volume is 1.184-1.145=0.039 cubic inches. The relative error is 0.039/1.184 ≈ 3.3%.
Comments
Post a Comment