Ever tried home-made potato chips? You can try doing some! All you need is a knife and a deep frier (and of course, a potato, salt, and oil). What characterizes potato chips from french fries is the shape of the cut. For a potato chip you need to do thin slices, so they can fry properly.
After slicing the potato you can begin to think backwards. Can you put all the slices back together to make the potato again? Maybe this is a little hard to do because, once sliced, you cannot reconstitute back the potato. However, the volume of the slices is equal to the volume of the potato before you sliced it.
The idea of slicing is very useful to find volumes because if the slices are very thin, you can take advantage of formulas for finding areas instead, and then reconstitute back the solid by means of integration.
What is the Meaning of Determining Volumes by Slicing?
You have probably found before formulas for finding the volumes of some geometric bodies, for example if you are given a cone, you can find its volume using the formula
\[ V_{\text{cone}} = \frac{1}{3}\pi r^2 h,\]
or maybe you are given a cylinder, whose volume can be found with
\[ V_{\text{cylinder}} = \pi r^2 h.\]
There is also the volume of a pyramid, which regardless of the shape of its base, you can find its volume with
\[ V_{\text{pyramid}} = \frac{1}{3}A h.\]
But where do all these formulas come from? Why are there \( \frac{1}{3} \) involved?
The answer to this question becomes easier if you slice any of these solids, obtaining a cross-section of the body.
A cross-section of a body is a view that shows how the body would look if you were to make a cut across it.
Consider the case of a cone. Here you can see a cross-section obtained by slicing the cone halfway through its height. The slice is done so it is parallel to the base of the cone.
Figure 1. The cross-section parallel to the base of a cone is a circle
This is particularly useful because the formula for finding the area of a circle is
\[ A_{\text{circle}} = \pi r^2, \]
so you can add together all the cross-sections of the cone by means of integration to obtain the volume of a cone.
Finding Volumes by Slicing
To find volumes by slicing, consider a simpler example: the volume of a rectangular prism.
Figure 2. A rectangular prism
Next, slice the prism to obtain a cross-section that is parallel to any of its faces, for example a cross-section parallel to its base.
Figure 3. The cross-section parallel to the base of a rectangular prism is a rectangle
You can find the area of a rectangle by using the formula
\[ A_{\text{rectangle}} = \ell w.\]
To find the volume of the rectangular prism, imagine you stack all the rectangular slices of the prism. Since all slices are equal, you just multiply the area of the rectangular prism (let's just call it \( A \) for simplicity) by the height of the prism, that is
\[ \begin{align} V_{\text{prism}} &= A h \\ &= \ell w h. \end{align} \]
You can obtain the same formula if stacking cross-sections that are parallel to any other side of the prism!
You can imagine this step as if dragging the area of a cross-section of the prism from one side to another
Figure 4. Volume of the rectangular prism obtained by dragging the area of a cross-section from one side to another
A similar idea applies to other geometric bodies, like cylinders.
Figure 5. Volume of a cylinder obtained by dragging the area of a cross-section from a base to another
The volume of a cylinder can be found by multiplying the area of the base by its height, so
\[ \begin{align} V_{\text{cylinder}} &= A h \\ &= \pi r^2 h. \end{align}\]
But what happens if the slices are not equal? You need to integrate!
Determining Formulas for Volumes by Slicing
So far you have seen that if all the areas of a cross-section of a solid are equal, you can just multiply by the length of the corresponding transversal side to obtain the volume of a solid, like in the case of the rectangular prism or with the cylinder.
Now consider the case of a cone. Suppose that the base of this cone is equal to \(R,\) and its height is \( h.\) If you take cross-sections parallel to its base, you will find out that these do not have the same area.
Figure 6. The cross-sections of a cone do not have the same area
For illustrative purposes, place the cone along the \(x-\)axis with its tip in the origin. This way you can find the radius of each cross-section as a function of \(x.\)
Figure 7. The radius of each cross-section is a function of \(x.\)
In fact, the cross-sections are all circles with different radii, which can be found by using similar triangles, that is
\[ \frac{R}{h} =\frac{r}{x},\]
so
\[ r = \frac{R}{h}x.\]
Figure 8. You can find the radius of a cross-section with the help of similar triangles
The function
\[ r(x) = \frac{R}{h}x\]
gives you the radius of each cross-section in terms of its position relative to the tip of the cone, so the area of a cross-section is
\[ \begin{align} A(x) &= \pi \left( r(x) \right) ^2 \\ &= \pi \left( \frac{R}{h}x \right)^2 \\ &= \frac{\pi R^2}{h^2}x^2. \end{align} \]
Now that you know a function for the area of each cross-section, you can add them together by means of integration. The integration limits are \(a=0\) since you are starting from the tip of the cone, and \( b=h\) because the cross-sections align all the way up to the base, so
\[ \begin{align} V &= \int_0^h A(x) \,\mathrm{d}x \\ &= \int_0^h \frac{\pi R^2}{h^2}x^2\,\mathrm{d}x \\ &= \frac{\pi R^2}{h^2}\int_0^h x^2\,\mathrm{d}x. \end{align}\]
You can solve the resulting definite integral using the Power Rule and the Fundamental Theorem of Calculus, that is
\[ \begin{align} \int_0^h x^2 \, \mathrm{d}x &= \left( \frac{1}{3}(h)^3 \right) - \left( \frac{1}{3}(0)^3\right) \\ &= \frac{1}{3}h^3. \end{align} \]
By plugging back the above definite integral into the formula for the volume, you obtain
\[ \begin{align} V &= \left( \frac{\pi R^2}{h^2} \right) \left(\frac{1}{3}h^3 \right) \\ &= \frac{\pi R^2h^3}{3h^2} \\ &= \frac{1}{3}\pi R^2 h,\end{align}\]
which is the formula for the volume of a cone,
\[V_{\text{cone}} = \frac{1}{3}\pi R^2h.\]
So that's where the \( \frac{1}{3} \) comes from! You can find formulas for more solids by following steps similar to above.
Determining Volumes by Slicing Solids of Revolution
Another way of finding volumes of solids is by rotation of two-dimensional figures around a revolution axis. The solids found this way are known as Solids of Revolution, and their volume are found using different methods depending on the solid.
Figure 9. A solid of revolution obtained by rotation of a parabola along the \(x-\)axis
Check out our articles about the Disk Method and the Washer Method for more information on this subject!
Examples of Volumes Determined by Slicing
There are more formulas that can be determined by slicing!
Find the formula for the volume of a square pyramid with side \( \ell \) and height \( h.\)
Answer:
The base of a square pyramid, as the name suggests, is square, so its area can be found using the formula for the area of a square, that is
\[ A_{\text{square}}=\ell ^2.\]
The cross-sections parallel to the base of the pyramid are smaller squares. You can use similar triangles to find that the length \( s \) of the sides of these squares are given by
\[ s(x) = \frac{\ell}{h}x. \]
Knowing this, the area of each cross-section becomes
\[ \begin{align} A(x) &= \left( s(x) \right)^2 \\ &= \left( \frac{\ell}{h}x\right)^2 \\ &= \frac{\ell^2}{h^2}x^2. \end{align} \]
By integrating this area along the \(x-\)axis from \( 0 \) to \( h, \) you can find the volume of the pyramid, so
\[ \begin{align} V &= \int_0^h A(x) \, \mathrm{d}x \\ &= \int_0^h \frac{\ell^2}{h^2} x^2 \, \mathrm{d}x \\ &= \frac{\ell^2}{h^2} \int_0^h x^2 \, \mathrm{d}x. \end{align}\]
The resulting definite integral is the same as the previous example of the cone, you can find it with the help of the Power Rule and the Fundamental Theorem of Calculus, giving you
\[ \int_0^h x^2 \, \mathrm{d}x = \frac{1}{3}h^3.\]
By plugging this integral back into the volume formula, you obtain
\[ \begin{align} V &= \left( \frac{\ell^2}{h^2} \right) \left( \frac{1}{3}h^3 \right) \\&= \frac{\ell^2h^3}{3h^2} \\ &= \frac{1}{3}\ell^2h. \end{align} \]
Note that the area of the base is \( A=\ell^2,\) so you can substitute this into the above formula and obtain
\[ V=\frac{1}{3}Ah,\]
which might look more familiar.
Ever wondered why the formula for the volume of a sphere has \( \frac{4}{3} \) in it?
Find the formula for the volume of a sphere with radius \( R.\)
Answer:
Begin by imagining you place the sphere in the origin of the coordinate system. The cross-sections of a sphere are all circles, so you can find the area of a cross-section of the sphere using
\[ A(x) = \pi \left(r(x)\right)^2.\]
You need to find the radius of the sphere depending on its position relative to the origin. In order to do so, focus on the cross-section of the sphere.
Fig. 10. The radius of a cross-section of the sphere with radius R can be expressed in terms of \(x\).
The radius of a cross-section satisfies the equation
\[x^2+r^2=R^2,\]
from where you can obtain
\[r^2=R^2-x^2,\]
so
\[\left( r(x) \right) ^2 = R^2-x^2.\]
This means that the area of a cross-section is given by
\[ A(x) = \pi (R^2-x^2),\]
so you can integrate to find the volume of the sphere, that is
\[ V(x) = \int_{-R}^R \pi \left(R^2-x^2\right) \, \mathrm{d}x\]
The integral can be done with the Power Rule. Do not forget that in this case, R is a constant!
\[ \begin{align} V(x) &= \int_{-R}^R \pi \left( R^2-x^2\right) \, \mathrm{d}x \\ &= \pi \left[ \left( R^2(R)-\frac{1}{3}(R)^3 \right) - \left( R^2(-R) -\frac{1}{3}(-R)^3 \right) \right] \\ &= \pi \left[ \left(R^3-\frac{1}{3}R^3 \right)-\left( -R^3+\frac{1}{3}R^3 \right)\right]\\ &=\pi \left[ \frac{2}{3}R^3+\frac{2}{3}R^3\right] \\ &= \frac{4}{3}\pi R^3. \end{align}\]
And that is how you obtain the formula for the volume of a sphere!
Determining Volumes by Slicing - Key takeaways
- A cross-section of a body is a view that show how the body would look if you were to make a cut across it.
- The idea behind determining volumes by slicing is to find the areas of the slices of a body, and integrate them to obtain their volume.
- In order to do so, you obtain slices, or cross-sections, aligned along the axis of integration.
- The formulas for many geometrical bodies can be found by slicing, like the volumes of pyramids, cones, and spheres.
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