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.
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 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 \(\De x=\frac{h}{n}\text{.}\) 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.
For each \(i=1,2,\cdots,n\text{,}\) pancake number \(i\) runs from \(x=x_{i-1}=(i-1)\cdot\De x\) to \(x=x_i=i\cdot\De x\text{,}\) 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 \(\De x\) and radius \(\frac{x_i^*}{h}r\text{.}\)
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:
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.
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 \(\De x = \frac{h}{n}\) 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 \De x && \text{which becomes}
\int_0^h \pi \left( \frac{x}{h} r \right)^2 \dee{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.
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
So the pancake at distance \(x\) from the \(yz\)-plane has
thickness 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 \(\De x=\frac{r}{n}\) and then take the limit \(n\rightarrow\infty\text{.}\) In the integral \(\De x\) is replaced by \(\dee{x}\text{.}\) Knowing that this is what is going to happen, we again just skip a few steps.
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.
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
Okay okay… We missed the pupil. I’m sure there is a pun in there somewhere.
. Here is what you see as the rotation takes place.
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.
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.
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 \(y=2\text{.}\)
Note that this is the full pancake, not just the part in the first octant.
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 \(\frac{h-0}{h}b=b\text{.}\)
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”.
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.
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.
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 \(R\text{!}\)
A \(45^\circ\) notch is cut to the centre 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.
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 centre 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
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
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).
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).
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.
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?
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.
Write down definite integrals that represent the following quantities. Do not evaluate the integrals explicitly.
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 \(0\le x\le 3\text{.}\)
Write down definite integrals that represent the following quantities. Do not evaluate the integrals explicitly.
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 \(y=-1\text{.}\)
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 \(x=5\text{.}\)
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.
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 \(\sqrt{\frac{2}{3}}\ell\text{.}\)
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 \(y=1\text{.}\)
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 \(\ln x\text{.}\))
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.
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.
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.
A frustum 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 frustum.
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).
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 \(a \geq b\text{.}\)
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 \(b\text{?}\)
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?
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.
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.
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.
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.
What is the mass of the atmosphere contained in a vertical column with radius one metre, sixty kilometres high?
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 \(\frac{x}{h}\cdot r\text{.}\)
