Calculating the volume of a cup via rotational solids
As part of tutoring physics and maths to high school students, I sometimes write up deep-dive explanations of questions arising during lessons. This post discusses two ways to find the volume of a cup generated by rotating two curves about the \(x\)-axis.
Question
A cup is generated by rotating two curves about the \(x\)-axis (see the blue region in the figure above). Which height, \(h\), must the cup have so that the rotational volume is approximately equal to \(200\ \text{cm}^3\)?
The functions are defined as:
\[f(x) = \sqrt{5x + 4}\]and
\[g(x) = \sqrt{5x - 4}.\]Answer
An important point to realise about this question is that we’re not trying to find the volume of the interior of the cup, i.e. that which would hold a fluid or another substance. The task is rather to find the volume of the material of the cup itself.
There are two ways to do this: (a) by finding the volume enclosed by each function and subtracting the smaller volume from the larger volume, and (b) by splitting the \(x\) domain into two well-defined regions and adding the solid formed by \(f(x)\) for the first region to the solid formed between \(f(x)\) and \(g(x)\) for the second region.
The first solution method is the easiest to understand, so we’ll start with that.
Subtracting rotational solids from one another
Before we begin solving this problem, there is an important point to make: notice that \(g(x)\) is not defined over the entire region that \(f(x)\) is defined. Where \(f(x)\) has a well-defined real value from \(x = 0\) and above, \(g(x)\) is not defined at \(x = 0\) and only obtains real values above a certain minimum \(x\) value. Consequently, it’s not possible to simply integrate from 0 up to \(h\) for both functions; we have to be careful.
To find the cup’s volume we need to calculate the volume enclosed by rotating \(g(x)\) about the \(x\)-axis and subtract this value from the volume enclosed by rotating \(f(x)\) about the \(x\)-axis.
Let’s calculate the volume bounded by \(f(x)\) first. We calculate the volume of the solid of revolution for \(f(x)\) by considering infinitesimally thin discs created by rotating \(f(x)\) around the \(x\)-axis and then integrating these discs over \(x\).
The surface area of a single disc is given by the familiar formula:
\[A = \pi r^2,\]where \(r\) is the distance from the \(x\)-axis to the function at a given \(x\) value. This distance is simply the function \(f(x)\), hence \(r = f(x)\). Thus, the area of a disc enclosed by the function \(f(x)\) at a given \(x\) is:
\[\begin{equation} A_f = \pi \bigl( f(x) \bigr)^2. \end{equation}\]Integrating these discs over \(x\) gives the volume enclosed by \(f(x)\), which we denote here by \(V_f\). I.e.
\[\begin{equation} V_f = \pi \int_a^b \bigl( f(x) \bigr)^2\ dx. \end{equation}\]We’re now in a position to carry out this integration. We know that
\[\begin{equation} f(x) = \sqrt{5x + 4}, \end{equation}\]hence we substitute this into our equation for \(V_f\) above, to get:
\[\begin{align} V_f &= \pi \int_a^b \Bigl( \sqrt{5x + 4} \Bigr)^2\ dx\\ &= \pi \int_a^b 5x + 4\ dx\\ &= \pi \biggl[\frac{5}{2}x^2 + 4x\biggr]_a^b\ dx. \end{align}\]To find the definite integral, we note that \(f(x)\) is bounded to the left by the \(y\)-axis (i.e. where \(x = 0\)) and to the right by the height of the cup, \(h\). Hence the integration bounds are \(a = 0\) and \(b = h\), which gives:
\[\begin{align} V_f &= \pi \biggl[\frac{5}{2}h^2 + 4h\biggr] - \pi \biggl[\frac{5}{2}0^2 + 0\biggr]\\ &= \pi \biggl[\frac{5}{2}h^2 + 4h\biggr]\\ &= \frac{5}{2} \pi h^2 + 4 \pi h. \end{align}\]To find the volume enclosed by \(g(x)\), we use a similar method. However, we note that the lower bound of the integral, \(a\), is no longer equal to 0; it is equal to the value of \(x\) when \(g(x) = 0\). The upper bound when integrating over \(g(x)\) is still \(b = h\).
We can calculate the lower bound by setting \(g(x)\) equal to zero and solving for \(x\):
\[\begin{alignat}{2} &&g(x) = 0 &= \sqrt{5x - 4};\ \text{(square both sides)}\\ \Rightarrow&& 0 &= 5 x - 4\\ \Rightarrow&& 5 x &= 4\\ \Rightarrow&& x &= \frac{4}{5}. \end{alignat}\]Hence the lower bound of the integral is \(a = \frac{4}{5}\).
The rotational volume of the solid enclosed by \(g(x)\) (denoted by \(V_g\)) can be written like so:
\[\begin{equation} V_g = \pi \int_a^b \bigl( g(x) \bigr)^2\ dx. \end{equation}\]Since we know that
\[\begin{equation} g(x) = \sqrt{5x - 4} \end{equation}\]we therefore have
\[\begin{align} V_g &= \pi \int_a^b 5x - 4\ dx\\ &= \pi \biggl[ \frac{5}{2} x^2 - 4x\biggr]_a^b. \end{align}\]Substituting \(a = \frac{4}{5}\) and \(b = h\), we get
\[\begin{align} V_g &= \pi \biggl[ \frac{5}{2} h^2 - 4h\biggr] - \pi \biggl[\frac{5}{2} \left(\frac{4}{5}\right)^2 - 4 \left(\frac{4}{5}\right)\biggr]\\ &= \pi \biggl[ \frac{5}{2} h^2 - 4h\biggr] - \pi \biggl[\frac{5}{2} \cdot \frac{16}{25} - \frac{16}{5}\biggr]\\ &= \pi \biggl[ \frac{5}{2} h^2 - 4h\biggr] - \pi \biggl[\frac{1}{2} \cdot \frac{16}{5} - \frac{16}{5}\biggr]\\ &= \pi \biggl[ \frac{5}{2} h^2 - 4h\biggr] - \pi \biggl[-\frac{1}{2} \cdot \frac{16}{5}\biggr]\\ &= \frac{5}{2} \pi h^2 - 4 \pi h + \frac{8}{5} \pi. \end{align}\]To find the volume of the cup, \(V_c\), we subtract the rotational volume enclosed by \(g(x)\) from the rotational volume enclosed by \(f(x)\):
\[\begin{align} V_c &= V_f - V_g\\ &= \frac{5}{2} \pi h^2 + 4 \pi h - \biggl(\frac{5}{2} \pi h^2 - 4 \pi h + \frac{8}{5} \pi\biggr)\\ &= \frac{5}{2} \pi h^2 + 4 \pi h - \frac{5}{2} \pi h^2 + 4 \pi h - \frac{8}{5} \pi\\ &= 8 \pi h - \frac{8}{5} \pi. \end{align}\]We want this volume to equal \(200\ \text{cm}^3\), hence we set \(V_c\) equal to 200 and solve for \(h\):
\[\begin{alignat}{2} &&V_c = 200 &= 8 \pi h - \frac{8}{5} \pi\\ \Rightarrow&& \frac{200}{8 \pi} &= h - \frac{1}{5}\\ \Rightarrow&& h &= \frac{200}{8 \pi} + \frac{1}{5}\\ \Rightarrow&& h &= \frac{25}{\pi} + \frac{1}{5}\ \text{cm}\\ \Rightarrow&& h &\approx 8.16\ \text{cm}. \end{alignat}\]Hence the desired height of the cup, \(h\), is approximately equal to 8.16 cm.
Integrating with discs and washers
As with many things in mathematics, there’s more than one way to do it. In the solution above, we subtracted the volume of the smaller solid from the volume of the larger solid to calculate the cup’s volume. Another way to approach the situation is to realise that we can split the \(x\) domain up into two regions: one where only \(f(x)\) is defined (\(x \in [0, 4/5)\)), and one where both \(f(x)\) and \(g(x)\) are defined (\(x \in [4/5, h]\)).
Looking at the problem from this perspective, we can find the solid of rotation for \(f(x)\) in the region where only \(f(x)\) is defined by using the method of discs. We can then find the solid of rotation between \(f(x)\) and \(g(x)\) over the domain where both \(f(x)\) and \(g(x)\) are defined by using the method of washers.
The following diagram illustrates the two regions over which we’ll be integrating, denoted by the text “discs” and “washers” respectively:
Adding the volumes of the rotational solids from each region will give us the same volume for the cup as calculated in the previous section. Let’s see how this works in detail.
As before, we have the functions
\[f(x) = \sqrt{5x + 4}\]and
\[g(x) = \sqrt{5x - 4}.\]First, we want to find the volume of the rotational solid formed by \(f(x)\) over the domain where \(g(x)\) isn’t defined, i.e. \(x \in [0, 4/5)\). As mentioned above, we use the method of discs to find the volume of this rotational solid, which we will denote here as \(V_{\text{discs}}\):
\[\begin{align} V_{\text{discs}} &= \pi \int_0^{4/5} \bigl( f(x) \bigr)^2\ dx\\ &= \pi \int_0^{4/5} \bigl( 5x + 4 \bigr)^2\ dx\\ &= \biggl[ \frac{5}{2} \pi x^2 + 4 \pi x \biggr]_0^{4/5}\\ &= \frac{5}{2} \biggl(\frac{4}{5}\biggr)^2 \pi + 4 \frac{4}{5} \pi\\ &= \frac{5}{2} \frac{16}{25} \pi + \frac{16}{5} \pi\\ &= \frac{1}{2} \frac{16}{5} \pi + \frac{16}{5} \pi\\ &= \frac{3}{2} \frac{16}{5} \pi\\ &= \frac{24}{5} \pi. \end{align}\]We find the rest of the cup’s volume by integrating the washers formed by the region between \(f(x)\) and \(g(x)\) when they are rotated about the \(x\)-axis. We then perform the integration from the minimum \(x\) value for which \(g(x)\) is defined up to \(h\) to get the volume for this part of the cup, which we will denote by \(V_{\text{washers}}\). This calculation is as follows:
\[\begin{align} V_{\text{washers}} &= \pi \int_{4/5}^h \bigl( f(x) \bigr)^2 - \bigl( g(x) \bigr)^2\ dx\\ &= \pi \int_{4/5}^h \bigl( 5x + 4 \bigr) - \bigl( 5x - 4 \bigr)\ dx\\ &= \pi \int_{4/5}^h 8\ dx\\ &= 8 \pi \biggl. x \biggr|_{4/5}^h\\ &= 8 \pi h - 8 \pi \frac{4}{5}. \end{align}\]To find the cup’s volume, we add \(V_\text{discs}\) to \(V_\text{washers}\):
\[\begin{align} V_\text{cup} &= V_\text{discs} + V_\text{washers}\\ &= \frac{24}{5} \pi + 8 \pi h - 8 \pi \frac{4}{5}. \end{align}\]We can extract a common factor of \(8 \pi\) by noting that \(24 = 3 \cdot 8\), thus
\[\begin{align} V_\text{cup} &= \frac{24}{5} \pi + 8 \pi h - 8 \pi \frac{4}{5}\\ &= 8 \pi \frac{3}{5} + 8 \pi h - 8 \pi \frac{4}{5}\\ &= 8 \pi \biggl(\frac{3}{5} + h - \frac{4}{5} \biggr)\\ &= 8 \pi \biggl(h - \frac{1}{5} \biggr). \end{align}\]We know that the desired cup volume should be equal to 200, hence we set this result equal to 200 and solve for \(h\):
\[\begin{align} V_\text{cup} &= 200 = 8 \pi \biggl(h - \frac{1}{5} \biggr)\\ \Rightarrow \frac{200}{8 \pi} &= h - \frac{1}{5}\\ \Rightarrow \frac{25}{\pi} &= h - \frac{1}{5}\\ \Rightarrow h &= \frac{25}{\pi} + \frac{1}{5}\ \text{cm}\\ \Rightarrow h &\approx 8.16\ \text{cm}. \end{align}\]which is the result we got from the previous section. Yay! 
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