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Volume of a Solid of Revolution: Washers

A solid of revolution is formed when an area in the plane is revolved around a line in the same plane. For example, Revolving a rectangle, produces a solid right circular cylinder. Revolving a semi-disk produces a solid spherical ball. The axis of revolution is the line around which the form is rotated.

There are two methods to solve this question: Disks and Washers.

This blog will introduce Washer Methods, which is well known as hollow soild revolution. In order to use washer methods, it could obtain two function and it could roated it either in \(x-axis\) or \(y-axis\). It is a demestraion of the solid roated around \(x-axis\).

For example, we want to calculate the volume of the solid obtained by rotating the region bounded by the parabola \(y=x^2\) and the square root function \(y=\sqrt{x}\) around the \(x-axis\)

In order to solve this problem, we want to know how it looks like. The following two graphs show that the two function in 2 dimentional, and 3 dementions after rotating the region bounded by those two functions aroundthe \(x-axis\).


From the graph above, we know those two function is bounded by 0 and 1. To approximate this solid by washer methods, we will cut the soild to part and calculate the volume of each part and add them together to get the solid volume approximation. In the function we choosed above, we will cut it to 10 parts, here is the demonstration in 3D model.

The graph above was 10 washer stack together, and they all have heigh equal to 1 one, and the outter radiams\((R_O)\) are the y value of function \(y=\sqrt{x}\), and the inner radiams \((R_I)\) are the y value of function \(y=x^2\), and the volume of those washers could be calculate by the function: \( V=\pi h (R_0^2-R_I^2)\). The following table show the volume of 10 washers and sum of their volume.
From the table above we got the approxamation of the solid is \(0.932 \ in^3 \). As we know in calculus, if we want to know accurate volume of a solid, we should use the Integral. For this question, from what we have been use above to calcuate the volue of each wahser, we should use the following integral to calculte the accurate volume: \[\pi \int_0^1 R_0(x)^2-R_I(x)^2 \ dx\] When we plug it into our question we got that: \[\begin{equation} V=\pi \int_{0}^{1}\left([\sqrt{x}]^{2}-\left[x^{2}\right]^{2}\right) d x=\pi \int_{0}^{1}\left(x-x^{4}\right) d x=\left.\pi\left(\frac{x^{2}}{2}-\frac{x^{5}}{5}\right)\right|_{0} ^{1}=\pi\left(\frac{1}{2}-\frac{1}{5}\right) =\frac{3 \pi}{10} \approx 0.942 in^3 \end{equation}\] The approxiamate value is really close to the accure value. However, if we cut the washer smaller and smaller, it will infenity close to the accurete value.

Author: Yueqi Li (Nicole)
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