There are many ways to “orient” the pyramid along the \(x\)-axis; Figure 7.2.4 gives one such way, with the pointed top of the pyramid at the origin and the \(x\)-axis going through the center of the base.

Each cross section of the pyramid is a square; this is a sample differential element. To determine its area \(A(x)\text{,}\) we need to determine the side lengths of the square.

When \(x=5\text{,}\) the square has side length 10; when \(x=0\text{,}\) the square has side length 0. Since the edges of the pyramid are lines, it is easy to figure that each cross-sectional square has side length \(2x\text{,}\) giving \(A(x) = (2x)^2=4x^2\text{.}\)

If one were to cut a slice out of the pyramid at \(x=3\text{,}\) as shown in Figure 7.2.5, one would have a shape with square bottom and top with sloped sides. If the slice were thin, both the bottom and top squares would have sides lengths of about 6, and thus the cross–sectional area of the bottom and top would be about 36in\(^2\text{.}\) Letting \(\Delta x_i\) represent the thickness of the slice, the volume of this slice would then be about \(36\Delta x_i\)in\(^3\text{.}\)

Cutting the pyramid into \(n\) slices divides the total volume into \(n\) equally–spaced smaller pieces, each with volume \((2x_i)^2\Delta x\text{,}\) where \(x_i\) is the approximate location of the slice along the \(x\)-axis and \(\Delta x\) represents the thickness of each slice. One can approximate total volume of the pyramid by summing up the volumes of these slices: \begin{equation*} \text{ Approximate volume } = \sum_{i=1}^n (2x_i)^2\Delta x. \end{equation*}

Taking the limit as \(n\to\infty\) gives the actual volume of the pyramid; recoginizing this sum as a Riemann Sum allows us to find the exact answer using a definite integral, matching the definite integral given by Theorem 7.2.2.

We have \begin{align*} V \amp = \lim_{n\to\infty} \sum_{i=1}^n (2x_i)^2\Delta x\\ \amp = \int_0^5 4x^2\ dx\\ \amp = \frac43x^3\Big|_0^5\\ \amp =\frac{500}{3}\ \text{ in } ^3 \approx 166.67\ \text{ in } ^3. \end{align*}

We can check our work by consulting the general equation for the volume of a pyramid (see the back cover under “Volume of A General Cone”):

\(\frac13\times \text{ area of base } \times \text{ height }\text{.}\)

Certainly, using this formula from geometry is faster than our new method, but the calculus–based method can be applied to much more than just cones.

A sketch can help us understand this problem. In Figure 7.2.8(a) the curve \(y=1/x\) is sketched along with the differential element — a disk — at \(x\) with radius \(R(x)=1/x\text{.}\) In Figure 7.2.8 (b) the whole solid is pictured, along with the differential element.

The volume of the differential element shown in part (a) of the figure is approximately \(\pi R(x_i)^2\Delta x\text{,}\) where \(R(x_i)\) is the radius of the disk shown and \(\Delta x\) is the thickness of that slice. The radius \(R(x_i)\) is the distance from the \(x\)-axis to the curve, hence \(R(x_i) = 1/x_i\text{.}\)

Slicing the solid into \(n\) equally–spaced slices, we can approximate the total volume by adding up the approximate volume of each slice: \begin{equation*} \text{ Approximate volume } = \sum_{i=1}^n \pi \left(\frac1{x_i}\right)^2\Delta x. \end{equation*}

Taking the limit of the above sum as \(n\to\infty\) gives the actual volume; recognizing this sum as a Riemann sum allows us to evaluate the limit with a definite integral, which matches the formula given in Key Idea 7.2.6: \begin{align*} V \amp = \lim_{n\to\infty}\sum_{i=1}^n \pi \left(\frac1{x_i}\right)^2\Delta x\\ \amp = \pi\int_1^2 \left(\frac1x\right)^2\ dx\\ \amp = \pi\int_1^2 \frac1{x^2}\ dx \end{align*} \begin{align*} \amp = \pi\left[-\frac1x\right]\Big|_1^2\\ \amp = \pi \left[-\frac12 - \left(-1\right)\right]\\ \amp = \frac{\pi}{2}\ \text{ units } ^3. \end{align*}

Since the axis of rotation is vertical, we need to convert the function into a function of \(y\) and convert the \(x\)-bounds to \(y\)-bounds. Since \(y=1/x\) defines the curve, we rewrite it as \(x=1/y\text{.}\) The bound \(x=1\) corresponds to the \(y\)-bound \(y=1\text{,}\) and the bound \(x=2\) corresponds to the \(y\)-bound \(y=1/2\text{.}\)

Thus we are rotating the curve \(x=1/y\text{,}\) from \(y=1/2\) to \(y=1\) about the \(y\)-axis to form a solid. The curve and sample differential element are sketched in Figure 7.2.12 (a), with a full sketch of the solid in Figure 7.2.12 (b).

We integrate to find the volume: \begin{align*} V \amp = \pi\int_{1/2}^1 \frac{1}{y^2}\ dy\\ \amp = -\frac{\pi}y\Big|_{1/2}^1\\ \amp = \pi\ \text{ units } ^3. \end{align*}

A sketch of the region will help, as given in Figure 7.2.19(a). Rotating about the \(x\)-axis will produce cross sections in the shape of washers, as shown in Figure 7.2.19(b); the complete solid is shown in part (c). The outside radius of this washer is \(R(x) = 2x+1\text{;}\) the inside radius is \(r(x) = x^2-2x+2\text{.}\) As the region is bounded from \(x=1\) to \(x=3\text{,}\) we integrate as follows to compute the volume. \begin{align*} V \amp = \pi\int_1^3 \Big((2x-1)^2-(x^2-2x+2)^2\Big)\ dx\\ \amp = \pi\int_1^3 \big(-x^4+4x^3-4x^2+4x-3\big)\ dx\\ \amp = \pi\Big[-\frac{1}{5}x^5+x^4-\frac43x^3+2x^2-3x\Big]\Big|_1^3\\ \amp =\frac{104}{15}\pi \approx 21.78\ \text{ units } ^3. \end{align*}

The triangular region is sketched in Figure 7.2.24(a); the differential element is sketched in (b) and the full solid is drawn in (c). They help us establish the outside and inside radii. Since the axis of rotation is vertical, each radius is a function of \(y\text{.}\)

The outside radius \(R(y)\) is formed by the line connecting \((2,1)\) and \((2,3)\text{;}\) it is a constant function, as regardless of the \(y\)-value the distance from the line to the axis of rotation is 2. Thus \(R(y)=2\text{.}\)

The inside radius is formed by the line connecting \((1,1)\) and \((2,3)\text{.}\) The equation of this line is \(y=2x-1\text{,}\) but we need to refer to it as a function of \(y\text{.}\) Solving for \(x\) gives \(r(y) = \frac12(y+1)\text{.}\)

We integrate over the \(y\)-bounds of \(y=1\) to \(y=3\text{.}\) Thus the volume is \begin{align*} V \amp = \pi\int_1^3\Big(2^2 - \big(\frac12(y+1)\big)^2\Big)\ dy\\ \amp = \pi\int_1^3\Big(-\frac14y^2-\frac12y+\frac{15}4\Big)\ dy\\ \amp = \pi\Big[-\frac1{12}y^3-\frac14y^2+\frac{15}4y\Big]\Big|_1^3\\ \amp = \frac{10}3\pi \approx 10.47\ \text{ units } ^3. \end{align*}