The volumes of many solids can be obtained by the application of the slice method. In the previous section, we sliced the solid into infinitely many thin disks. However, the element of volume does not have to be necessarily a disk (or a washer).

In a more general case, imagine the solid is sliced into infinitesimally thin slices of thickness \(dx\) by a family of planes perpendicular to the \(x\)-axis (Figure 1). Suppose you know a formula for the area of an arbitrary cross-section of the solid made by such planes \(A(x)\). Some common cross-sections are triangles, squares, rectangles, trapezoids, and semicircles. Then the volume of a slice (the element of volume) is this area \(A(x)\) multiplied by the thickness of the slide \(dx\): \[dV=A(x)dx.\] The total volume of the solid is the sum of the volumes of these slices. If the solid is bounded by two parallel planes perpendicular to the \(x\)-axis at \(x=a\) and \(x=b\), then \[\boxed{V=\int_{a}^{b}dV=\int_{a}^{b}A(x)dx.}\]

Figure 1

Similarly, suppose that the solid is cut by planes perpendicular to the \(y\)-axis and \(A(y)\) is the area of an arbitrary cross section (Figure 2). Then \[\boxed{V=\int_{c}^{d}A(y)dy}\] if the solid is bounded between the planes \(y=c\) and \(y=d\) (\(c<d\)).

Figure 2

Example 1
Compute the volume of a sphere with radius \(R\), knowing that the area of a circle with radius \(r\) is \(\pi r^{2}\).
Solution 1
Let’s cut the sphere by planes perpendicular to the \(x\)-axis into thin slices with circular cross section area \(A(x)\) (Figure 3).

Figure 3

If the distance of the plane from the origin is \(x\), then the radius of the cross section is \(\sqrt{R^{2}-x^{2}}\). Therefore, the area of the cross section is \[A(x)=\pi\left(R^{2}-x^{2}\right)\] and the volume of the slice is \[dV=A(x)dx=\pi\left(R^{2}-x^{2}\right)dx\] Because \(x\) varies between \(-R\) and \(R\), the volume of the sphere is \[\begin{aligned} V & =\int_{-R}^{R}\pi\left(R^{2}-x^{2}\right)dx\\ & =2\int_{0}^{R}\pi\left(R^{2}-x^{2}\right)dx\\ & =2\pi\left[R^{2}x-\frac{x^{3}}{3}\right]_{0}^{R}\\ & =\frac{4\pi}{3}R^{3}.\end{aligned}\]

Example 2
Consider two circular cylinders of radius \(a\), which intersect at right angles as shown in Figure 4. Compute the volume of the region common to both cylinders.

Figure 4

Solution 2
Take the axes of the cylinders to be the \(y\)– and \(z\)-axes. Figure 5 shows one-eighth of the common region. Let’s cut this solid \(ABCDO\) by planes perpendicular to the \(x-\)axis into thin slices. The area of the cross section \(EFGH\) is \[A(x)=EF\times EH\] and from the symmetry \(EF=EH\). Therefore, \[A(x)=EF^{2}.\]

(a) (b)

Figure 5

 

If \(OF=x\), because \(OE=a\), we have \[EF=\sqrt{a^{2}-x^{2}}.\] Therefore, \[A(x)=a^{2}-x^{2}\] and the volume of this slice is \[dV=A(x)dx=\left(a^{2}-x^{2}\right)dx\] The total volume of the common region is \[\begin{aligned} V & =8\int_{0}^{a}dV=8\int_{0}^{a}\left(a^{2}-x^{2}\right)dx\\ & =8\left[a^{2}x-\frac{x^{3}}{3}\right]_{0}^{a}\\ & =8\times\frac{2a^{3}}{3}\\ & =\frac{16a^{3}}{3}.\end{aligned}\]

Example 3
Consider a right circular cylinder of radius \(a\). Find the volume of the wedge cut from this cylinder by a plane through a diameter of the base of the cylinder making a \(45^{\circ}\) angle with the base.
Solution 3
Method (a): Taking the axes as shown in Figure 6, let the wedge be divided into thin slices of thickness $dx$ by planes perpendicular to the $x$-axis. Figure 6 shows one of these slices. As we can see, the sections cut by the planes are right triangles. 

 

(a) (b)

Figure 6

If the slice is located at a distance $x$ from the origin, the base of the triangle is \(b=\sqrt{a^{2}-x^{2}}\) and its height is \(h=b\tan45^{\circ}=\sqrt{a^{2}-x^{2}}\). Thus \[A(x)=\frac{1}{2}bh=\frac{1}{2}\left(a^{2}-x^{2}\right)\] and the volume of the infinitely thin slice is \[dV=A(x)dx=\frac{1}{2}\left(a^{2}-x^{2}\right)dx\] Because \(x\) varies between \(-a\) and \(a\) the total volume is \[\begin{aligned} \int_{-a}^{a}dV & =\int_{-a}^{a}\frac{1}{2}\left(a^{2}-x^{2}\right)dx\\ & =2\int_{0}^{a}\frac{1}{2}\left(a^{2}-x^{2}\right)dx\\ & =\left[a^{2}x-\frac{1}{3}x^{3}\right]_{0}^{a}\\ & =\frac{2a^{3}}{3}.\end{aligned}\]

Method (b): Imagine that the wedge is cut up into thin slices of thickness $dy$ by planes perpendicular to the $y$-axis. Figure 7 shows one of these slices. As we can see the slice has a rectangular face. 

(a) (b)

Figure 7

If the slice is at a distance $y$ from the origin,  the width of the rectangle is \(2\sqrt{a^{2}-y^{2}}\), and its height is \(y\tan 45^\circ=y\). Therefore, the area of the rectangle is \[A(y)=2y\sqrt{a^{2}-y^{2}},\] and the volume of the slice is \[dV=A(y)dy=2y\sqrt{a^{2}-y^{2}}\] Because \(y\) varies between \(0\) and \(a\), the total volume of the wedge is: \[V=\int_{0}^{a}dV=\int_{0}^{a}2y\sqrt{a^{2}-y^{2}}dy\] To evaluate the above integral, let \(u=a^{2}-y^{2}\). Then \[du=-2ydy\] \[y=0\Leftrightarrow u=a^{2}\] \[y=a\Leftrightarrow u=0\] and \[\begin{aligned} V & =\int_{0}^{a}2y\sqrt{a^{2}-y^{2}}\,dy\\ & =\int_{0}^{a}\underbrace{\sqrt{a^{2}-y^{2}}}_{\sqrt{u}}\underbrace{2y\,dy}_{-du}\\ & =-\int_{a^{2}}^{0}\sqrt{u}\,du\\ & =\int_{0}^{a^{2}}\sqrt{u}\,du\\ & =\frac{2}{3}\left.u^{\frac{3}{2}}\right|_{0}^{a^{2}}\\ & =\frac{2}{3}a^{3},\end{aligned}\] as before.