# 1.6: Volumes

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Another simple 1 application of integration is computing volumes. We use the same strategy as we used to express areas of regions in two dimensions as integrals — approximate the region by a union of small, simple pieces whose volume we can compute and then then take the limit as the “piece size” tends to zero.

In many cases this will lead to “multivariable integrals” that are beyond our present scope 2. But there are some special cases in which this leads to integrals that we can handle. Here are some examples.

##### Example 1.6.1 Cone

Find the volume of the circular cone of height $$h$$ and radius $$r\text{.}$$

Solution

Here is a sketch of the cone.

We have called the vertical axis $$x\text{,}$$ just so that we end up with a “$$\, d{x}$$” integral.

• In what follows we will slice the cone into thin horizontal “pancakes”. In order to approximate the volume of those slices, we need to know the radius of the cone at a height $$x$$ above its point. Consider the cross sections shown in the following figure.

At full height $$h\text{,}$$ the cone has radius $$r\text{.}$$ If we cut the cone at height $$x\text{,}$$ then by similar triangles (see the figure on the right) the radius will be $$\frac{x}{h}\cdot r\text{.}$$

• Now think of cutting the cone into $$n$$ thin horizontal “pancakes”. Each such pancake is approximately a squat cylinder of height $$\Delta x=\frac{h}{n}\text{.}$$ This is very similar to how we approximated the area under a curve by $$n$$ tall thin rectangles. Just as we approximated the area under the curve by summing these rectangles, we can approximate the volume of the cone by summing the volumes of these cylinders. Here is a side view of the cone and one of the cylinders.
• We follow the method we used in Example 1.5.1, except that our slices are now pancakes instead of rectangles.
• Pick a natural number $$n$$ (that we will later send to infinity), then
• subdivide the cone into $$n$$ thin pancakes, each of width $$\Delta x=\frac{h}{n}\text{.}$$
• For each $$i=1,2,\cdots,n\text{,}$$ pancake number $$i$$ runs from $$x=x_{i-1}=(i-1)\cdot\Delta x$$ to $$x=x_i=i\cdot\Delta x\text{,}$$ and we approximate its volume by the volume of a squat cone. We pick a number $$x_i^*$$ between $$x_{i-1}$$ and $$x_i$$ and approximate the pancake by a cylinder of height $$\Delta x$$ and radius $$\frac{x_i^*}{h}r\text{.}$$
• Thus the volume of pancake $$i$$ is approximately $$\pi \left( \frac{x_i^*}{h}r\right)^2 \Delta x$$ (as shown in the figure above).
• So the Riemann sum approximation of the volume is

\begin{align*} \text{Area} &\approx \sum_{i=1}^n \pi \left( \frac{x_i^*}{h}r\right)^2 \Delta x \end{align*}

• By taking the limit as $$n \to \infty$$ (i.e. taking the limit as the thickness of the pancakes goes to zero), we convert the Riemann sum into a definite integral (see Definition 1.1.9) and at the same time our approximation of the volume becomes the exact volume:

\begin{gather*} \int_0^h \pi \Big(\frac{x}{h}r\Big)^2\, d{x} \end{gather*}

Our life 3 would be easier if we could avoid all this formal work with Riemann sums every time we encounter a new volume. So before we compute the above integral, let us redo the above calculation in a less formal manner.

• Start again from the picture of the cone

and think of slicing it into thin pancakes, each of width $$\, d{x}\text{.}$$

• The pancake at height $$x$$ above the point of the cone (which is the fraction $$\frac{x}{h}$$ of the total height of the cone) has

As $$x$$ runs from $$0$$ to $$h\text{,}$$ the total volume is

\begin{align*} \int_0^h \pi \Big(\frac{x}{h}r\Big)^2\, d{x} &=\frac{\pi r^2}{h^2}\int_0^h x^2\, d{x}\\ &=\frac{\pi r^2}{h^2} \bigg[\frac{x^3}{3}\bigg]_0^h\\ &=\frac{1}{3}\pi r^2 h \end{align*}

• radius $$\frac{x}{h}\cdot r$$ (the fraction $$\frac{x}{h}$$ of the full radius, $$r$$) and so
• cross-sectional area $$\pi \big(\frac{x}{h}r\big)^2\text{,}$$
• thickness $$\, d{x}$$ — we have done something a little sneaky here, see the discussion below.
• volume $$\pi \big(\frac{x}{h}r\big)^2\, d{x}$$

In this second computation we are using a time-saving trick. As we saw in the formal computation above, what we really need to do is pick a natural number $$n\text{,}$$ slice the cone into $$n$$ pancakes each of thickness $$\Delta x = \frac{h}{n}$$ and then take the limit as $$n \to \infty\text{.}$$ This led to the Riemann sum

\begin{align*} \sum_{i=1}^n \pi \left( \frac{x_i^*}{h} r \right)^2 \Delta x && \text{which becomes} \int_0^h \pi \left( \frac{x}{h} r \right)^2 \, d{x} \end{align*}

So knowing that we will replace

\begin{align*} \sum_{i=1}^n &\longrightarrow \int_0^h\\ x_i^* &\longrightarrow x\\ \Delta x &\longrightarrow \, d{x} \end{align*}

when we take the limit, we have just skipped the intermediate steps. While this is not entirely rigorous, it can be made so, and does save us a lot of algebra.

##### Example 1.6.2 Sphere

Find the volume of the sphere of radius $$r\text{.}$$

Solution

We'll find the volume of the part of the sphere in the first octant 4, sketched below. Then we'll multiply by $$8\text{.}$$

• To compute the volume,

we slice it up into thin vertical “pancakes” (just as we did in the previous example).

• Each pancake is one quarter of a thin circular disk. The pancake a distance $$x$$ from the $$yz$$-plane is shown in the sketch above. The radius of that pancake is the distance from the dot shown in the figure to the $$x$$-axis, i.e. the $$y$$-coordinate of the dot. To get the coordinates of the dot, observe that
• it lies the $$xy$$-plane, and so has $$z$$-coordinate zero, and that
• it also lies on the sphere, so that its coordinates obey $$x^2+y^2+z^2=r^2\text{.}$$ Since $$z=0$$ and $$y \gt 0\text{,}$$ $$y=\sqrt{r^2-x^2}\text{.}$$
• So the pancake at distance $$x$$ from the $$yz$$-plane has
• thickness 5 $$dx$$ and
• radius $$\sqrt{r^2-x^2}$$
• cross-sectional area $$\frac{1}{4}\pi \big(\sqrt{r^2-x^2}\,\big)^2$$ and hence
• volume $$\frac{\pi}{4} \big(r^2-x^2\big)\, d{x}$$
• As $$x$$ runs from $$0$$ to $$r\text{,}$$ the total volume of the part of the sphere in the first octant is

\begin{gather*} \int_0^r \frac{\pi}{4} \big(r^2-x^2\big)\, d{x} =\frac{\pi}{4}\bigg[r^2x-\frac{x^3}{3}\bigg]_0^r =\frac{1}{6}\pi r^3 \end{gather*}

and the total volume of the whole sphere is eight times that, which is $$\frac{4}{3}\pi r^3\text{,}$$ as expected.
##### Example 1.6.3 Revolving a region

The region between the lines $$y=3\text{,}$$ $$y=5\text{,}$$ $$x=0$$ and $$x=4$$ is rotated around the line $$y=2\text{.}$$ Find the volume of the region swept out.

Solution

As with most of these problems, we should start by sketching the problem.

• Consider the region and slice it into thin vertical strips of width $$\, d{x}\text{.}$$
• Now we are to rotate this region about the line $$y=2\text{.}$$ Imagine looking straight down the axis of rotation, $$y=2\text{,}$$ end on. The symbol in the figure above just to the right of the end the line $$y=2$$ is supposed to represent your eye 6. Here is what you see as the rotation takes place.
• Upon rotation about the line $$y=2$$ our strip sweeps out a “washer”
• whose cross-section is a disk of radius $$5-2=3$$ from which a disk of radius $$3-2=1$$ has been removed so that it has a
• cross-sectional area of $$\pi 3^2 -\pi 1^2 = 8\pi$$ and a
• thickness $$\, d{x}$$ and hence a
• volume $$8\pi\,\, d{x}\text{.}$$
• As our leftmost strip is at $$x=0$$ and our rightmost strip is at $$x=4\text{,}$$ the total

\begin{align*} \text{Volume} &= \int _0^4 8\pi\,\, d{x} =(8\pi)(4) =32\pi \end{align*}

Notice that we could also reach this answer by writing the volume as the difference of two cylinders.

• The outer cylinder has radius $$(5-2)$$ and length 4. This has volume

\begin{align*} V_{outer} &= \pi r^2 \ell = \pi \cdot 3^2 \cdot 4 = 36\pi. \end{align*}

• The inner cylinder has radius $$(3-2)$$ and length 4. This has volume

\begin{align*} V_{inner} &= \pi r^2 \ell = \pi \cdot 1^2 \cdot 4 = 4\pi. \end{align*}

• The volume we want is the difference of these two, namely

\begin{align*} V &= V_{outer} - V_{inner} = 32\pi. \end{align*}

Let us turn up the difficulty a little on this last example.

##### Example 1.6.4 Revolving again

The region between the curve $$y=\sqrt{x}\text{,}$$ and the lines $$y=0\text{,}$$ $$x=0$$ and $$x=4$$ is rotated around the line $$y=0\text{.}$$ Find the volume of the region swept out.

Solution

We can approach this in much the same way as the previous example.

• Consider the region and cut it into thin vertical strips of width $$\, d{x}\text{.}$$
• When we rotate the region about the line $$y=0\text{,}$$ each strip sweeps out a thin pancake
• whose cross-section is a disk of radius $$\sqrt{x}$$ with a
• cross-sectional area of $$\pi (\sqrt{x})^2 = \pi x$$ and a
• thickness $$\, d{x}$$ and hence a
• volume $$\pi x \, d{x}\text{.}$$
• As our leftmost strip is at $$x=0$$ and our rightmost strip is at $$x=4\text{,}$$ the total

\begin{align*} \text{Volume} &= \int _0^4 \pi x \, d{x} =\left[\frac{\pi}{2}x^2 \right]_0^4 =8\pi \end{align*}

In the last example we considered rotating a region around the $$x$$-axis. Let us do the same but rotating around the $$y$$-axis.

##### Example 1.6.5 Revolving yet again

The region between the curve $$y=\sqrt{x}\text{,}$$ and the lines $$y=0\text{,}$$ $$x=0$$ and $$x=4$$ is rotated around the line $$x=0\text{.}$$ Find the volume of the region swept out.

Solution:

• We will cut the region into horizontal slices, so we should write $$x$$ as a function of $$y\text{.}$$ That is, the region is bounded by $$x=y^2\text{,}$$ $$x=4\text{,}$$ $$y=0$$ and $$y=2\text{.}$$
• Now slice the region into thin horizontal strips of width $$\, d{y}\text{.}$$
• When we rotate the region about the line $$x=0\text{,}$$ each strip sweeps out a thin washer
• whose inner radius is $$y^2$$ and outer radius is $$4\text{,}$$ and
• thickness is $$\, d{y}$$ and hence
• has volume $$\pi(r_{out}^2 - r_{in}^2)\, d{y} = \pi(16-y^4)\, d{y}\text{.}$$
• As our bottommost strip is at $$y=0$$ and our topmost strip is at $$y=2\text{,}$$ the total

\begin{align*} \text{Volume} &= \int _0^2 \pi(16-y^4) \, d{y} =\left[16\pi y - \frac{\pi}{5}y^5 \right]_0^2 = 32\pi - \frac{32\pi}{5} = \frac{128\pi}{5}. \end{align*}

There is another way 7 to do this one which we show at the end of this section.

##### Example 1.6.6 Pyramid

Find the volume of the pyramid which has height $$h$$ and whose base is a square of side $$b\text{.}$$

Solution

Here is a sketch of the part of the pyramid that is in the first octant; we display only this portion to make the diagrams simpler.

Note that this diagram shows only 1 quarter of the whole pyramid.

• To compute its volume, we slice it up into thin horizontal “square pancakes”. A typical pancake also appears in the sketch above.
• The pancake at height $$z$$ is the fraction $$\frac{h-z}{h}$$ of the distance from the peak of the pyramid to its base.
• So the full pancake 8 at height $$z$$ is a square of side $$\frac{h-z}{h}b\text{.}$$ As a check, note that when $$z=h$$ the pancake has side $$\frac{h-h}{h}b=0\text{,}$$ and when $$z=0$$ the pancake has side $$\frac{h-0}{h}b=b\text{.}$$
• So the pancake has cross-sectional area $$\big(\frac{h-z}{h}b\big)^2$$ and thickness 9 $$\, d{z}$$ and hence
• volume $$\big(\frac{h-z}{h}b\big)^2\, d{z}\text{.}$$
• The volume of the whole pyramid (not just the part of the pyramid in the first octant) is \begin{align*} \int_0^h \Big(\frac{h-z}{h}b\Big)^2\, d{z} &=\frac{b^2}{h^2} \int_0^h (h-z)^2\, d{z}\\ \end{align*}

Now use the substitution rule with $$t=(h-z), \, d{z}\to-\, d{t}$$

\begin{align*} &=\frac{b^2}{h^2} \int_h^0 -t^2\, d{t}\\ &=-\frac{b^2}{h^2}\bigg[\frac{t^3}{3}\bigg]_h^0\\ &=-\frac{b^2}{h^2}\bigg[-\frac{h^3}{3}\bigg]\\ &=\frac{1}{3} b^2h \end{align*}

Let's ramp up the difficulty a little.

##### Example 1.6.7 Napkin Ring

Suppose you make two napkin rings 10 by drilling holes with different diameters through two wooden balls. One ball has radius $$r$$ and the other radius $$R$$ with $$r \lt R\text{.}$$ You choose the diameter of the holes so that both napkin rings have the same height, $$2h\text{.}$$ See the figure below.

Which 11 ring has more wood in it?

Solution

We'll compute the volume of the napkin ring with radius $$R\text{.}$$ We can then obtain the volume of the napkin ring of radius $$r\text{,}$$ by just replacing $$R \mapsto r$$ in the result.

• To compute the volume of the napkin ring of radius $$R\text{,}$$ we slice it up into thin horizontal “pancakes”. Here is a sketch of the part of the napkin ring in the first octant showing a typical pancake.
• The coordinates of the two points marked in the $$yz$$-plane of that figure are found by remembering that
• the equation of the sphere is $$x^2+y^2+z^2=R^2\text{.}$$
• The two points have $$y \gt 0$$ and are in the $$yz$$-plane, so that $$x=0$$ for them. So $$y=\sqrt{R^2-z^2}\text{.}$$
• In particular, at the top of the napkin ring $$z=h\text{,}$$ so that $$y=\sqrt{R^2-h^2}\text{.}$$
• The pancake at height $$z\text{,}$$ shown in the sketch, is a “washer” — a circular disk with a circular hole cut in its center.
• The outer radius of the washer is $$\sqrt{R^2-z^2}$$ and
• the inner radius of the washer is $$\sqrt{R^2-h^2}\text{.}$$ So the
• cross-sectional area of the washer is

\begin{gather*} \pi\big(\sqrt{R^2-z^2}\,\big)^2-\pi\big(\sqrt{R^2-h^2}\,\big)^2 =\pi(h^2-z^2) \end{gather*}

• The pancake at height $$z$$
• has thickness $$dz$$ and
• cross-sectional area $$\pi(h^2-z^2)$$ and hence
• volume $$\pi(h^2-z^2)\, d{z}\text{.}$$
• Since $$z$$ runs from $$-h$$ to $$+h\text{,}$$ the total volume of wood in the napkin ring of radius $$R$$ is

\begin{align*} \int_{-h}^h \pi(h^2-z^2)\, d{z} &=\pi\Big[h^2z-\frac{z^3}{3}\Big]_{-h}^h\\ &=\pi\Big[\Big(h^3-\frac{h^3}{3}\Big) -\Big((-h)^3-\frac{(-h)^3}{3}\Big)\Big]\\ &=\pi\Big[\frac{2}{3}h^3-\frac{2}{3}\big(-h\big)^3\Big]\\ &=\frac{4\pi}{3}h^3 \end{align*}

This volume is independent of $$R\text{.}$$ Hence the napkin ring of radius $$r$$ contains precisely the same volume of wood as the napkin ring of radius $$R\text{!}$$

##### Example 1.6.8 Notch

A $$45^\circ$$ notch is cut to the center of a cylindrical log having radius $$20$$cm. One plane face of the notch is perpendicular to the axis of the log. See the sketch below. What volume of wood was removed?

Solution

We show two solutions to this problem which are of comparable difficulty. The difference lies in the shape of the pancakes we use to slice up the volume. In solution 1 we cut rectangular pancakes parallel to the $$yz$$-plane and in solution 2 we slice triangular pancakes parallel to the $$xz$$-plane.

Solution 1:

• Concentrate on the notch. Rotate it around so that the plane face lies in the $$xy$$-plane.
• Then slice the notch into vertical rectangles (parallel to the $$yz$$-plane) as in the figure on the left below.

• The cylindrical log had radius $$20$$cm. So the circular part of the boundary of the base of the notch has equation $$x^2+y^2=20^2\text{.}$$ (We're putting the origin of the $$xy$$-plane at the center of the circle.) If our coordinate system is such that $$x$$ is constant on each slice, then
• the base of the slice is the line segment from $$(x,-y,0)$$ to $$(x,+y,0)$$ where $$y=\sqrt{20^2-x^2}$$ so that
• the slice has width $$2y=2\sqrt{20^2-x^2}$$ and
• height $$x$$ (since the upper face of the notch is at $$45^\circ$$ to the base — see the side view sketched in the figure on the right above).
• So the slice has cross-sectional area $$2x\sqrt{20^2-x^2}\text{.}$$
• On the base of the notch $$x$$ runs from $$0$$ to $$20$$ so the volume of the notch is \begin{align*} V&=\int_0^{20}2x\sqrt{20^2-x^2}\, d{x}\\ \end{align*}

Make the change of variables $$u=20^2-x^2$$ (don't forget to change $$\, d{x} \rightarrow -\frac{1}{2x}\, d{u}$$):

\begin{align*} V&=\int_{20^2}^{0}-\sqrt{u}\,\, d{u}\\ &=\left[-\frac{u^{3/2}}{3/2}\right]_{20^2}^0\\ &= \frac{2}{3}20^3=\frac{16,000}{3} \end{align*}

Solution 2:

• Concentrate of the notch. Rotate it around so that its base lies in the $$xy$$-plane with the skinny edge along the $$y$$-axis.
• Slice the notch into triangles parallel to the $$xz$$-plane as in the figure on the left below. In the figure below, the triangle happens to lie in a plane where $$y$$ is negative.

• The cylindrical log had radius $$20$$cm. So the circular part of the boundary of the base of the notch has equation $$x^2+y^2=20^2\text{.}$$ Our coordinate system is such that $$y$$ is constant on each slice, so that
• the base of the triangle is the line segment from $$(0,y,0)$$ to $$(x,y,0)$$ where $$x=\sqrt{20^2-y^2}$$ so that
• the triangle has base $$x=\sqrt{20^2-y^2}$$ and
• height $$x=\sqrt{20^2-y^2}$$ (since the upper face of the notch is at $$45^\circ$$ to the base — see the side view sketched in the figure on the right above).
• So the slice has cross-sectional area $$\frac{1}{2}\big(\sqrt{20^2-y^2}\big)^2\text{.}$$
• On the base of the notch $$y$$ runs from $$-20$$ to $$20\text{,}$$ so the volume of the notch is

\begin{align*} V&=\frac{1}{2}\int_{-20}^{20}(20^2-y^2)\, d{y}\\ &=\int_0^{20} (20^2-y^2)\, d{y}\\ &=\Big[20^2y-\frac{y^3}{3}\Big]_0^{20}\\ &=\frac{2}{3}20^3=\frac{16,000}{3} \end{align*}

## Optional — Cylindrical shells

Let us return to Example 1.6.5 in which we rotate a region around the $$y$$-axis. Here we show another solution to this problem which is obtained by slicing the region into vertical strips. When rotated about the $$y$$-axis, each such strip sweeps out a thin cylindrical shell. Hence the name of this approach (and this subsection).

##### Example 1.6.9 Revolving yet again

The region between the curve $$y=\sqrt{x}\text{,}$$ and the lines $$y=0\text{,}$$ $$x=0$$ and $$x=4$$ is rotated around the line $$x=0\text{.}$$ Find the volume of the region swept out.

Solution

• Consider the region and cut it into thin vertical strips of width $$\, d{x}\text{.}$$
• When we rotate the region about the line $$y=0\text{,}$$ each strip sweeps out a thin cylindrical shell
• whose radius is $$x\text{,}$$
• height is $$\sqrt{x}\text{,}$$ and
• thickness is $$\, d{x}$$ and hence
• has volume $$2 \pi \times \text{radius} \times \text{height} \times \text{thickness} = 2 \pi x^{3/2} \, d{x}\text{.}$$
• As our leftmost strip is at $$x=0$$ and our rightmost strip is at $$x=4\text{,}$$ the total

\begin{align*} \text{Volume} &= \int _0^4 2\pi x^{3/2} \, d{x} =\left[\frac{4\pi}{5} x^{5/2} \right]_0^4 = \frac{4\pi}{5}\cdot 32 = \frac{128\pi}{5} \end{align*}

which (thankfully) agrees with our previous computation.

## Exercises

##### 1

Consider a right circular cone.

What shape are horizontal cross-sections? Are the vertical cross-sections the same?

##### 2

Two potters start with a block of clay $$h$$ units tall, and identical square cookie cutters. They form columns by pushing the square cookie cutter straight down over the clay, so that its cross-section is the same square as the cookie cutter. Potter A pushes their cookie cutter down while their clay block is sitting motionless on a table; Potter B pushes their cookie cutter down while their clay block is rotating on a potter's wheel, so their column looks twisted. Which column has greater volume?

##### 3

Let $$R$$ be the region bounded above by the graph of $$y=f(x)$$ shown below and bounded below by the $$x$$-axis, from $$x=0$$ to $$x=6\text{.}$$ Sketch the washers that are formed by rotating $$R$$ about the $$y$$-axis. In your sketch, label all the radii in terms of $$y\text{,}$$ and label the thickness.

##### 4 (✳)

Write down definite integrals that represent the following quantities. Do not evaluate the integrals explicitly.

1. The volume of the solid obtained by rotating around the $$x$$--axis the region between the $$x$$--axis and $$y=\sqrt{x}\, e^{x^2}$$ for $$0\le x\le 3\text{.}$$
2. The volume of the solid obtained by revolving the region bounded by the curves $$y=x^2$$ and $$y=x+2$$ about the line $$x=3\text{.}$$
##### 5 (✳)

Write down definite integrals that represent the following quantities. Do not evaluate the integrals explicitly.

1. The volume of the solid obtained by rotating the finite plane region bounded by the curves $$y=1-x^2$$ and $$y=4-4x^2$$ about the line $$y=-1\text{.}$$
2. The volume of the solid obtained by rotating the finite plane region bounded by the curve $$y=x^2-1$$ and the line $$y=0$$ about the line $$x=5\text{.}$$
##### 6 (✳)

Write down a definite integral that represents the volume of the solid obtained by rotating around the line $$y=-1$$ the region between the curves $$y=x^2$$ and $$y=8-x^2\text{.}$$ Do not evaluate the integrals explicitly.

##### 7

A tetrahedron is a three-dimensional shape with four faces, each of which is an equilateral triangle. (You might have seen this shape as a 4-sided die; think of a pyramid with a triangular base.) Using the methods from this section, calculate the volume of a tetrahedron with side-length $$\ell\text{.}$$ You may assume without proof that the height of a tetrahedron with side-length $$\ell$$ is $$\sqrt{\frac{2}{3}}\ell\text{.}$$

##### 8 (✳)

Let $$a \gt 0$$ be a constant. Let $$R$$ be the finite region bounded by the graph of $$y=1+\sqrt{x}e^{x^2}\text{,}$$ the line $$y=1\text{,}$$ and the line $$x=a\text{.}$$ Using vertical slices, find the volume generated when $$R$$ is rotated about the line $$y=1\text{.}$$

##### 9 (✳)

Find the volume of the solid generated by rotating the finite region bounded by $$y = 1/x$$ and $$3x + 3y = 10$$ about the $$x$$--axis.

##### 10 (✳)

Let $$R$$ be the region inside the circle $$x^2 + (y-2)^2=1\text{.}$$ Let $$S$$ be the solid obtained by rotating $$R$$ about the $$x$$-axis.

1. Write down an integral representing the volume of $$S\text{.}$$
2. Evaluate the integral you wrote down in part (a).
##### 11 (✳)

The region $$R$$ is the portion of the first quadrant which is below the parabola $$y^2=8x$$ and above the hyperbola $$y^2-x^2=15\text{.}$$

1. Sketch the region $$R\text{.}$$
2. Find the volume of the solid obtained by revolving $$R$$ about the $$x$$ axis.
##### 12 (✳)

The region $$R$$ is bounded by $$y=\log x\text{,}$$ $$y=0\text{,}$$ $$x=1$$ and $$x=2\text{.}$$ (Recall that we are using $$\log x$$ to denote the logarithm of $$x$$ with base $$e\text{.}$$ In other courses it is often denoted $$\ln x\text{.}$$)

1. Sketch the region $$R\text{.}$$
2. Find the volume of the solid obtained by revolving this region about the $$y$$ axis.
##### 13 (✳)

The finite region between the curves $$y = \cos(\frac x2)$$ and $$y = x^2 - \pi^2$$ is rotated about the line $$y=-{\pi^2}\text{.}$$ Using vertical slices (disks and/or washers), find the volume of the resulting solid.

##### 14 (✳)

The solid $$V$$ is 2 meters high and has square horizontal cross sections. The length of the side of the square cross section at height $$x$$ meters above the base is $$\frac{2}{1+x}$$ m. Find the volume of this solid.

##### 15 (✳)

Consider a solid whose base is the finite portion of the $$xy$$--plane bounded by the curves $$y=x^2$$ and $$y=8-x^2\text{.}$$ The cross--sections perpendicular to the $$x$$--axis are squares with one side in the $$xy$$--plane. Compute the volume of this solid.

##### 16 (✳)

A frustrum of a right circular cone (as shown below) has height $$h\text{.}$$ Its base is a circular disc with radius $$4$$ and its top is a circular disc with radius $$2\text{.}$$ Calculate the volume of the frustrum.

##### 17

The shape of the earth is often approximated by an oblate spheroid, rather than a sphere. An oblate spheroid is formed by rotating an ellipse about its minor axis (its shortest diameter).

1. Find the volume of the oblate spheroid obtained by rotating the upper (positive) half of the ellipse $$(ax)^2+(by)^2=1$$ about the $$x$$-axis, where $$a$$ and $$b$$ are positive constants with $$a \geq b\text{.}$$
2. Suppose 12 the earth has radius at the equator of 6378.137 km, and radius at the poles of 6356.752 km. If we model the earth as an oblate spheroid formed by rotating the upper half of the ellipse $$(ax)^2+(by)^2=1$$ about the $$x$$-axis, what are $$a$$ and $$b\text{?}$$
3. What is the volume of this model of the earth? (Use a calculator.)
4. Suppose we had calculated the volume of the earth by modelling it as a sphere with radius $$6378.137$$ km. What would our absolute and relative errors be, compared to our oblate spheroid calculation?
##### 18 (✳)

Let $$R$$ be the bounded region that lies between the curve $$y = 4 - (x - 1)^2$$ and the line $$y = x + 1\text{.}$$

1. Sketch $$R$$ and find its area.
2. Write down a definite integral giving the volume of the region obtained by rotating $$R$$ about the line $$y = 5\text{.}$$ Do not evaluate this integral.
##### 19 (✳)

Let $$\cR=\big\{(x,y)\ :\ (x-1)^2+y^2\le 1\text{ and } x^2+(y-1)^2\le 1\ \big\}\text{.}$$

1. Sketch $$\cR$$ and find its area.
2. If $$\cR$$ rotates around the $$y$$--axis, what volume is generated?
##### 20 (✳)

Let $$\cR$$ be the plane region bounded by $$x=0,\ x=1,\ y=0$$ and $$y=c\sqrt{1+x^2}\text{,}$$ where $$c$$ is a positive constant.

1. Find the volume $$V_1$$ of the solid obtained by revolving $$\cR$$ about the $$x$$--axis.
2. Find the volume $$V_2$$ of the solid obtained by revolving $$\cR$$ about the $$y$$--axis.
3. If $$V_1=V_2\text{,}$$ what is the value of $$c\text{?}$$
##### 21 (✳)

The graph below shows the region between $$y = 4 + \pi \sin x$$ and $$y = 4 + 2\pi - 2x\text{.}$$

The region is rotated about the line $$y = -1\text{.}$$ Express in terms of definite integrals the volume of the resulting solid. Do not evaluate the integrals.

##### 22

On a particular, highly homogeneous 13 planet, we observe that the density of the atmosphere $$h$$ kilometres above the surface is given by the equation $$\rho(h) = c2^{-h/6}\quad \frac{\mathrm{kg}}{\mathrm{m^3}}\text{,}$$ where $$c$$ is the density on the planet's surface.

1. What is the mass of the atmosphere contained in a vertical column with radius one metre, sixty kilometres high?
2. What height should a column be to contain $$\dfrac{3000c\pi}{\log 2}$$ kilograms of air?
1. Well — arguably the idea isn't too complicated and is a continuation of the idea used to compute areas in the previous section. In practice this can be quite tricky as we shall see.
2. Typically such integrals (and more) are covered in a third calculus course.
3. At least the bits of it involving integrals.
4. The first octant is the set of all points $$(x,y,z)$$ with $$x\ge 0\text{,}$$ $$y\ge 0$$ and $$z\ge 0\text{.}$$
5. Yet again what we really do is pick a natural number $$n\text{,}$$ slice the octant of the sphere into $$n$$ pancakes each of thickness $$\Delta x=\frac{r}{n}$$ and then take the limit $$n\rightarrow\infty\text{.}$$ In the integral $$\Delta x$$ is replaced by $$dx\text{.}$$ Knowing that this is what is going to happen, we again just skip a few steps.
6. Okay okay… We missed the pupil. I'm sure there is a pun in there somewhere.
7. The method is not a core part of the course and should be considered optional.
8. Note that this is the full pancake, not just the part in the first octant.
9. We are again using our Riemann sum avoiding trick.
10. Handy things to have (when combined with cloth napkins) if your parents are coming to dinner and you want to convince them that you are “taking care of yourself”.
11. A good question to ask to distract your parents from the fact you are serving frozen burritos.
12. Earth Fact Sheet, NASA, accessed 2 July 2017
13. This is clearly a simplified model: air density changes all the time, and depends on lots of complicated factors aside from altitude. However, the equation we're using is not so far off from an idealized model of the earth's atmosphere, taken from Pressure and the Gas Laws by H.P. Schmid, accessed 3 July 2017.

This page titled 1.6: Volumes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.