6.5 Slicing Method II

6.5.1 Selecting a Convenient Scanning Axis

The slicing method can be applied in contexts where (i) the solid may not have a base, or (ii) even though the solid has a basis, it may be more convenient to slice the solid along a different axis other than by slicing it along a rectangular approximation of the base. The following two examples illustrate these two cases

Example 1.
A spherical reservoir of radius 2 m contains water 3 m deep. Find the volume of water in the reservoir.

Figure 6.17a

Figure 6.17b

Figure 6.17c

Solution. Figure 6.17 shows

1. the reservoir and the water in it;
2. a slicing of the water in the reservoir. The slicing is done along the vertical \(z\)-axis as, in this case, the cross sections are circles whose area can be readily computed as \(\pi r^2\), for a cross section of radius \(r\).
3. this is a representation of the vertical cross section of the reservoir passing through the center together with the scanning \(z\)-axis. Note that we have placed the center of the reservoir at the origin.

Obsevations. Note the following in connection with Figure 6.17(c),

1. A horizontal cross section of the water corresponds to a horizontal line located at the coordinate \(z\).
2. The radius of the cross section has been denoted by \(r\).
3. The origin of the \(z\)-axis is at the center of the reservoir. This is convenient as the Pythagorean theorem can be easily applied to find the radius \(r\) of the cross sections in terms of \(z\).
4. Because of the choice of origin, the interval for the depth in the water is given by \(z\in[-2,1]\), while the actual depth for a point located at \(z\)-level is given by depth \( d = z+2\), \(d \in [0,3]\)..
5. Then, \(\displaystyle r = \sqrt{2^2-z^2}\), \(z\in[-2,1]\) (explain why).

Step 1. Partition of the interval.

\[z\in[-2,1], \;\;\; \Delta z = \frac{1-(-2)}{n}=\frac{3}{n}, \;\;\; z_i=-2+i\Delta z=-2 + \frac{3i}{n}, \; i=0,1,2,\dots , n.\]

Step 2. Volumes of the slices.

\[V_i=A(z_i)\Delta z\;\;\; \mathrm{and} \;\;\; A(z)=\pi r^2 =\pi \left(\sqrt{2^2-z_i^2}\right)^2.\]

Step 3. Riemann sum obtained by adding the volumes of the slices.

\[\sum _{i=1}^n V_i
=\sum _{i=1}^n A\left(z_i\right)\Delta z
=\sum _{i=1}^n \pi\left(\sqrt{2^2-z_i^2}\right)^2 \Delta z
=\sum _{i=1}^n \pi\left(4-z_i^2\right) \Delta z.\]

Step 4. Evaluation of the definite integral.

It is possible to compute the limit of the Riemann sums, note that
\begin{align*} \sum _{i=1}^n A\left(z_i\right)\Delta z &= \sum _{i=1}^n \pi\left(4-z_i^2\right) \Delta z=\sum _{i=1}^n \pi\left(4-\left(-2 + \frac{3i}{n}\right)^2\right) \frac{3}{n} = \\
&\underset{?}{=}\sum _{i=1}^n \pi \left(\frac{12i}{n}-\frac{9i^2}{n^2}\right)\frac{3}{n} \underset{?}{=}\pi \left(\frac{12}{n} \frac{n(n+1)}{2} -\frac{9}{n^2}\frac{n(n+1)(2n+1)}{6} \right)\frac{3}{n}.
\end{align*}
\begin{align*}
\int_{-2}^{1}A(z)\,dz
&=\lim_{n\to \infty} \sum _{i=1}^n A\left(z_i\right)\,\Delta z
=\lim_{n\to \infty} \pi \left(\frac{12}{n} \frac{n(n+1)}{2} -\frac{9}{n^2}\frac{n(n+1)(2n+1)}{6} \right)\frac{3}{n}
\underset{?}{=}9\pi
,\end{align*}
or
\[\int_{-2}^{1}\pi \left(\sqrt{4-z^2}\right)^2\,dz
=\lim_{n\to \infty} \sum _{i=1}^n \pi \left(\sqrt{4-z_i^2}\right)^2\Delta z =9\pi.\]
Alternatively, we can apply TFTCP2:
\begin{align*}
\int_{-2}^{1}A(z)\,dz
&=\int_{-2}^{1}\pi \left(\sqrt{4-z^2}\right)^2\,dz=\int_{-2}^{1}\pi \left(4-z^2\right)\,dz
=\pi \left(4z-\frac{1}{3} z^3\right)\,\Big|_{-2}^{1}=\\
& =\pi \left (\left(4-\frac{1}{3} \right)-\left(-8-\frac{-8}{3}\right)\right ) = 9\pi.
\end{align*}

Example 2.
Show that a pyramid with a square base of sides \(a\), and height \(h\) has volume \(\displaystyle V=\frac{1}{3}a^2h\).

Figure 6.18a

Figure 6.18b

Figure 6.18c

Solution. Figure 6.18 shows

1. the pyramid;
2. a slicing of the pyramid. The slicing is done along the vertical \(z\)-axis as the cross sections in this case are squares, whose area can be readily computed as \(s^2\), for a cross section of with sides of length \(s\).
3. a vertical cross section of the pyramid (passing through the apex) together with the scanning \(z\)-axis.

Obsevations. Note the following in connection with Figure 6.18(c),

1. A horizontal cross section of the pyramid corresponds to a horizontal line located at height \(z\).
2. The length of each side of the cross section is denoted by \(s\).
3. The origin of the \(z\)-axis is at the base of the pyramid.
4. Because of the choice of origin, the interval for the \(z\)-coordinate is given by \(z\in[0,h]\).
5. Then, using similar triangles, \(\displaystyle s = \frac{a}{h}(h-z)\), \(z\in[0,h]\) (explain why).

Step 1. Partition of the interval.

\[z\in[0, h], \;\;\; \Delta z = \frac{h}{n}, \;\;\; z_i=-i\Delta z= \frac{h i}{n}, \; i=0,1,2,\dots , n.\]

Step 2. Volumes of the slices.

\[V_i=A(z_i)\Delta z\;\;\; \mathrm{and}\;\;\; (A(z)=s^2 =\left(\frac{a}{h}(h-z_i)\right)^2.\]

Step 3. Riemann sum obtained by adding the volumes of the slices.

\begin{align*}
\sum _{i=1}^n V_i
&=\sum _{i=1}^n A\left(z_i\right)\Delta z
=\sum _{i=1}^n \left(\frac{a}{h}(h-z_i)\right)^2 \Delta z.
\end{align*}

Step 4. Evaluation of the definite integral.

We can compute the limit of the Riemann sums, or, we can use TFTCP2.

To compute the limit, note that
\begin{align*}
\sum _{i=1}^n \left(\frac{a}{h}(h-z_i)\right)^2 \Delta z&=\sum _{i=1}^n \frac{a^2}{h^2}\left(h^2-2hz_i + z_i^2\right) \Delta z =\sum _{i=1}^n \frac{a^2}{h^2}\left(h^2 – 2h \frac{h i}{n} + \left(\frac{h i}{n}\right)^2\right) \frac{h}{n}=\\
& \underset{?}{=} \frac{a^2}{hn}
\left(h^2 n -\frac{2h^2}{n} \frac{n(n+1)}{2} +\frac{h^2}{n^2}\frac{n(n+1)(2n+1)}{6} \right).
\end{align*}
\begin{align*}
\int_{0}^{h}A(z)\,dz
&=\lim_{n\to \infty} \sum _{i=1}^n A\left(z_i\right)\,\Delta z
=\sum _{i=1}^n \frac{a^2}{h^2}\left(h^2-2hz_i + z_i^2\right) \Delta z =\\
&=\lim_{n\to \infty} \frac{a^2}{hn}
\left(h^2 n -\frac{2h^2}{n} \frac{n(n+1)}{2} +\frac{h^2}{n^2}\frac{n(n+1)(2n+1)}{6} \right)
\underset{?}{=}a^2\frac{h}{3}=\frac{1}{3}a^2h.
\end{align*}
\begin{align*}
\int_{0}^{h}A(z)\,dz =\int_{0}^{h}\frac{a^2}{h^2}\left(h^2-2hz + z^2\right)\, dz=
\frac{a^2}{h^2}\left(h^2z-hz^2 +\frac{1}{3} z^3\right)\,\Big|_{0}^{h}=\frac{1}{3}a^2h.
\end{align*}