6.2 Slicing Method I

6.2.1 Visualization of the Slicing Method

&Nbsp;

Example 1
Let \(\mathcal{S}\) be the solid whose base is the region \(\mathcal{R}\) enclosed by the graph of the ellipse
\(x^2+4y^2=36\),
and whose cross sections perpendicular to the \(x\)-axis are equilateral triangles. Find the volume of the solid.

Figure 6.4 shows the following:

1. The solid.
2. The solid together with a right endpoint rectangular approximation to the region \(\mathcal{R}\), and the corresponding cross sections of the solid at the right-endpoints of the partition.

Note. The cross sections are obtained by scanning the solid along the \(x\)-axis because of the condition: cross sections perpendicular to the \(x\)-axis.

3. The corresponding approximating prisms to the slices constructed with the cross sections and the rectangles in the rectangular approximation.

Figure 6.4a

Figure 6.4b

Figure 6.4c

The slicing method consists in adding the volumes of the prisms thus obtained to set up a Riemann sum, and then taking the limit as the number of prisms goes to infinity, to evaluate the corresponding definite integral. To accomplish this, we need first to translate to this context the formula to compute the volume of a prism.

6.2.2 Prior Knowledge and Reformulation in Context

• The volume of a right prism with base of area \(A\) and height \(h\) is given by \(V= A h\).
• In the context of the prisms approximating the slices of the solid,
  let \(A(x_i)\) denote the area of the cross section at \(x_i\). Note that \(A(x_i)\) plays the role of the base of the prism, and that \(\Delta x\) is precisely the height/thickness of the prism. Then, the volume of the \(i\)-th prism corresponding to the \(i\)-th rectangular region is given by
  \begin{equation}\label{ch06-02-eq01}\tag{6.4}
  V_i= A(x_i)\Delta x,
  \end{equation}

6.2.3 Riemann Sum and Definite Integral

Adding the volumes of the approximating \(n\) prisms yields a Riemann sum,
\begin{equation}\label{ch06-02-eq02}\tag{6.5}
\sum _{i=1}^n V_i=\sum _{i=1}^n A\left(x_i\right)\Delta x.
\end{equation}
and, the limit of the Riemann sums as \(n\to \infty\) yields a definite integral.
\begin{equation}\label{ch06-02-eq03}\tag{6.6}
V(\mathcal{S}) = \int_a^bA(x)\,dx = \lim_{n\to \infty}\sum _{i=1}^n A\left(x_i\right)\Delta x.
\end{equation}

6.2.4 Method and Example

Example 1 and Method
(Continuation.) Let \(\mathcal{S}\) be the solid whose base is the region \(\mathcal{R}\) enclosed by the graph of the ellipse \(x^2+4y^2=36\),
and whose cross sections perpendicular to the \(x\)-axis are equilateral triangles. Find the volume of
\(\mathcal{S}\).

Figure 6.5a

Figure 6.5b

Solution.

Step 1. Rectangular approximation to the region $\mathcal{R}$.

Partition of the interval:
\(\displaystyle \Delta x=\frac{6-(-6)}{n} =\frac{12}{n}, \;\;\; x_i=x_0+i \Delta x=-6+\frac{12i}{n}\).

Width of the rectangular regions:
\(\displaystyle w_i=\Delta x =\frac{12}{n}\).

Height of the \(i\)-th rectangular region.

1. Think of the graph of the ellipse \(x^2+4y^2=36\) as consisting of the graph of two functions \(f\) and \(g\) as shown in Figure 6.5.
   \[f(x)=\frac{1}{2}\sqrt{36-x^2}, \;\;\; \text{ and }\;\;\; g(x)=-\frac{1}{2}\sqrt{36-x^2}.\]
   The expressions for \(f\) and \(g\) are obtained by solving for \(y\) in \(x^2+4y^2=36\).
1. The part above the \(x\)-axis is precisely the graph of \(f\).
2. The part below the \(x\)-axis is precisely the graph of \(g\).
2. The height of the \(i\)-th rectangular region is given by the distance between the \(y\)-coordinates of the points \((x_i,f(x_i))\) and \((x_i,g(x_i))\) as can be seen in Figure 6.5. Namely,
   \[h_i=f\left(x_i\right) – g\left(x_i\right)=\frac{1}{2}\sqrt{36-x_i^2}-\left(-\frac{1}{2}\sqrt{36-x_i^2}\right)
   =\sqrt{36-x_i^2}.\]

Step 2. Volume of the prisms corresponding to the rectangular regions.

Since the volume of the \(i\)-th prism is given by \(V_i=A(x_i)\Delta x\) (see (\ref{ch06-02-eq01})), we need to compute \(A(x_i)\).

• (Prior Knowledge.) An equilateral triangle with sides of length \(s\) has area
  \(\displaystyle A=\frac{\sqrt{3}}{4}s^2\). (Use the Pythagorean Theorem.)
• The base of the cross section at the right-end-point \(x_i\) is given by \(f(x_i)-g(x_i)\), that is, the base is the height of the \(i\)-th rectangular region, \(h_i\).
• It follows that the area of the cross section \(A(x_i)\) is given by
  \[A(x_i)=\frac{\sqrt{3}}{4}(h_i)^2 =\frac{\sqrt{3}}{4}(f(x_i)-g(x_i))^2,\;\;\; i=1,2,\dots, n.\]

Step 3. Riemann sum to approximate the volume.

\[
\sum _{i=1}^n V_i
=\sum _{i=1}^n A\left(x_i\right)\Delta x
=\sum _{i=1}^n \frac{\sqrt{3}}{4}\left(f(x_i)-g(x_i)\right)^2\Delta x
=\sum _{i=1}^n \frac{\sqrt{3}}{4}\left(\sqrt{36-x_i^2}\right)^2\Delta x.\]
\[ =\sum _{i=1}^n \frac{\sqrt{3}}{4}\left(36-x_i^2\right)\Delta x\]
\[ =\sum _{i=1}^n \frac{\sqrt{3}}{4}\left(36-x_i^2\right)\Delta x=\sum _{i=1}^n \frac{\sqrt{3}}{4}\left(36-\left(-6+\frac{12i}{n}\right)^2\right)\frac{12}{n} \underset{?}{=}\sum _{i=1}^n \frac{\sqrt{3}}{4}\left(\frac{144i}{n}-\frac{144i^2}{n^2}\right)\frac{12}{n}=\]
\[\underset{?}{=} \frac{\sqrt{3}}{4}\frac{12}{n}\left(\frac{144}{n}\sum _{i=1}^n i-\frac{144}{n^2}\sum _{i=1}^n i^2\right) \underset{?}{=} \frac{\sqrt{3}}{4}\frac{12}{n}\left( \frac{144}{n}\frac{n(n+1)}{2} -\frac{144}{n^2}\frac{n(n+1)(2n+1)}{6} \right).\]

Step 4. Evaluation of the definite integral to compute the volume.

\[\int_{-6}^{6}A(x)\,dx = \int_{-6}^{6}\frac{\sqrt{3}}{4}\left(f(x)-g(x)\right)^2\,dx
= \int_{-6}^{6}\frac{\sqrt{3}}{4}\left(36-x^2\right) \,dx.\]
To evaluate this definite integral, we can compute the limit
\begin{align*}
\int_{-6}^{6}A(x)\,dx
&=\lim_{n\to \infty} \sum _{i=1}^n A\left(x_i\right)\,\Delta x
=\lim_{n\to \infty} \sum _{i=1}^n \frac{\sqrt{3}}{4}\left(f(x_i)-g(x_i)\right)^2\Delta x
\underset{?}{=} 72\sqrt{3},
\end{align*}
\begin{align*}
&=\lim_{n\to \infty}\frac{\sqrt{3}}{4}\frac{12}{n}\left(\frac{144}{n}\frac{n(n+1)}{2} -\frac{144}{n^2}\frac{n(n+1)(2n+1)}{6}\right) \underset{?}{=} 72\sqrt{3} .\end{align*}
or, we can apply TFTCP2
\[\int_{-6}^{6}\frac{\sqrt{3}}{4}\left(f(x)-g(x)\right)^2\,dx
= \int_{-6}^{6}\frac{\sqrt{3}}{4}\left(36-x^2\right) \,dx
=\frac{\sqrt{3}}{4} \left(36x-\frac{1}{3}x^3\right)\,\Big|_{-6}^{6}
=72\sqrt{3}
.\]

On Your Own 1.

Let \(\mathcal{S}\) be the solid whose base is the region \(\mathcal{R}\) enclosed by the graphs of
\(y=\sqrt{x}\), \(x=1\), and \(y=0\),
and whose cross sections perpendicular to the \(x\)-axis are semicircles. Find the volume of \(\mathcal{S}\). (See Figure 6.6.)

Figure 6.6a

Figure 6.6b

6.2.5 Scanning along the $y$-axis

Example 1
Let \(\mathcal{S}\) be the solid whose base is the region \(\mathcal{R}\) enclosed by the graphs of the parabola \(y = 4x-x^2\) and the line \(y=0\),
and whose cross sections perpendicular to the \(y\)-axis are isosceles right triangles whose hypotenuse is parallel to the \(x\)-axis (see Figure 6.7(a)).
Set up the limit of Riemann sum and its representation as a definite integral to compute the volume of the solid.

Figure 6.7a

Figure 6.7b

Figure 6.7c

On Your Own 2.
Let \(\mathcal{S}\) be a solid as in Example 1, but with cross sections
(a) squares; (b) semicircles. Set up the limit of Riemann sums and its representation as a definite integral to compute the volume of the solid.

Solution. First of all, note that the phrase “cross sections perpendicular to the \(y\)-axis” carries the connotation that the solid is to be scanned/sliced along the \(y\)-axis. Since the three solids share the same base, Step 1 is the same for the three of them.

Because the slicing has to be done along the \(y\)-axis, we need to think of the parabolas as consisting of the graph of two functions \(f\) and \(g\) that depend on \(y\), as shown in Figure 6.8(a). To accomplish this, we solve for \(x\) in \(y=4x-x^2\).

1. The vertex of the parabola is at \((2,4)\), thus, the functions are defined for \(y \in [0,4]\).
2. The part on the left of the parabola corresponds to the graph of \(g(y)=2-\sqrt{4-y}\).
3. The part on the right of the parabola corresponds to the the graph of \(f(y)=2+\sqrt{4-y}\).

Figure 6.8a

Figure 6.8b

Step 1. Rectangular approximation to the region

Partition of the interval:
\(\displaystyle \Delta y=\frac{4-0}{n} =\frac{4}{n}, \;\;\; y_i=y_0+i \Delta y=\frac{4i}{n}\).
Width of the rectangular regions:
\(\displaystyle w_i=\Delta y=\frac{4}{n}\).
Length of the \(i\)-th rectangular region.
The length of the \(i\)-th rectangular region is given by the distance between the \(x\)-coordinates of the points \((f(y_i),y_i)\) and \((g(y_i),y_i)\) as can be seen in Figure 6.8(b). Namely,
\[l_i=f\left(y_i\right) – g\left(y_i\right)
=2+\sqrt{4-y_i} – \left(2-\sqrt{4-y_i}\right) = 2\sqrt{4-y_i}
.\]

Step 2. Volume of the prisms corresponding to the rectangular regions.

Since the volume of the \(i\)-th prism is given by \(V_i=A(y_i)\Delta y\), we need to find a formula for
\(A(y_i)\).

• (Prior Knowledge.) For an isosceles right triangle with legs of length \(s\), its area is given by
  \(\displaystyle A=\frac{1}{2}s^2\) (explain why).
• The base of the cross section at the right-end-point \(y_i\) is given by \(f(y_i)-g(y_i)\), that is, the base is the length of the \(i\)-th rectangular region, \(l_i\).
• Since the base is the hypotenuse, and the legs have the same length, say \(s\), then we have
  \[s^2+s^2=l_i^2\Longleftrightarrow s=\frac{1}{\sqrt{2}}l_i
  \Longrightarrow s=
  \frac{1}{\sqrt{2}}(f(y_i)-g(y_i))=
  \frac{1}{\sqrt{2}}\left(2\sqrt{4-y_i}\right)=\sqrt{2}\sqrt{4-y_i}.\]
• It follows that the area of the cross section \(A(y_i)\) is given by
  \[A(y_i)=\frac{1}{2}(s)^2 =\frac{1}{2}\left(\frac{1}{\sqrt{2}}(f(y_i)-g(y_i))\right)^2
  =\frac{1}{2}\left(\sqrt{2}\sqrt{4-y_i}\right)^2 = 4-y_i,\;\;\; i=1,2,\dots, n.\]

Step 3. Riemann sum to approximate the volume.

We can now set up the Riemann sum to approximate the volume of the solid by adding the volumes of the prisms.
\begin{align*}
\sum _{i=1}^n V_i
&=\sum _{i=1}^n A\left(y_i\right)\Delta y
=\sum _{i=1}^n \frac{1}{2}\left(\frac{1}{\sqrt{2}}(f(y_i)-g(y_i))\right)^2 \Delta y
=\sum _{i=1}^n (4-y_i)\Delta y
=\sum _{i=1}^n \left(4-\frac{4i}{n}\right)\frac{4}{n}
.
\end{align*}

Step 4. Evaluation of the definite integral to compute the volume.

\[Vol(\mathcal{S})=
\int_{0}^{4}A(y)\,dy
=\lim_{n\to \infty} \sum _{i=1}^n A\left(y_i\right)\,\Delta y ,\]
\[\int_{0}^{4}\left(4-y\right)\,dy
=\lim_{n\to \infty} \sum _{i=1}^n \left(4-y_i\right)\Delta y = \lim_{n\to \infty} \sum _{i=1}^n \left(4-\frac{4i}{n}\right)\frac{4}{n}\underset{?}{=}8.\]
Alternatively, by TFTCP2, we have
\[\int_{0}^{4}\left(4-y\right)\,dy = \left(4y-\frac{1}{2} y^2\right)\,\Big|_{0}^{4} = 8.\]