9 Week 9

In this section, we consider the mean and root mean square values of a function, which is important in studying sinusoidal type functions.

9.1 Mean and RMS Values

The mean value for a function \(y = f(x)\) for the range \(a\) to \(b\) is the mean value of all the ordinates in that range. Consider a function \(y = f(x)\) for the values \(x = a\) to \(x = b\), divided into a number of strips of equal width \(\delta x\). Let the mid-ordinate for each strip be \(y_1, y_2, y_3,\ldots,y_n\), see Figure 9.1.

Figure 9.1: \(f(x)\)

The mean or average value of the mid-ordinates is \[ \overline y=\frac{y_1+\ldots+y_n}{n} \]

Multiplying the numerator and denominator by \(\delta x\) gives \[ \overline y=\frac{y_1\delta x+\ldots+y_n\delta x}{n\delta x} \] and since \(n\delta x = b - a\) we have \[ \overline y = \frac{\sum y\delta x}{b-a}. \] When the number of strips increases indefinitely such that \(\delta x \to 0\) \[ \overline y = \frac{1}{b-a}\int_a^b y\ \mathrm{d}x. \] That is, the mean value of a function between two limits is the area under the curve between those limits divided by the range of \(x\) between those limits.

Example 9.1 If the velocity \(v\) in \(m/s\) of a body is given by \(v=4t+3\) where \(t\) is in seconds, determine the mean velocity of the body from \(t=2\ \)s to \(t=6\ \)s.

Solution. \[\begin{align*} \text{Mean velocity}& = \frac{1}{6-2}\int_2^6(4t+3)\ \mathrm{d}t = \frac 14\left[2t^2+3t\right]_2^6\\ &= \frac 14\left[(2\times36+3\times6) -(2\times4+3\times2)\right] = \frac{76}{4}=19\ m/s. \end{align*}\]

9.1.1 RMS Value

The RMS (or root mean square) value is obtained by taking \(f(x)\), squaring it, finding the mean value and then taking the square root.

Its importance in engineering is demonstrated when a sinusoidal function is investigated, for example when the average value of the current or voltage in a circuit is to be found. For all sine or cosine functions, the mean value over a complete cycle is zero because the curve is equally distributed above and below the \(x-\)axis. However, if the ordinates are squared first, there are then no negative ordinates and an average may be found. Ammeters and voltmeters for a.c. circuits are usually calibrated to find the RMS value.

To summarise, the RMS value is defined to be the number \[ \text{RMS} = \sqrt{\frac{1}{b-a}\int_a^b y^2\ \mathrm{d}x}. \]

Example 9.2 Determine the RMS value of the current in an a.c. circuit given by \(i=20+100\sin(100\pi t)\) between \(t=0\) and \(t=0.02\) s.

Solution. \[\begin{align*} \text{RMS}&= \sqrt{\frac{1}{b-a}\int_a^b i^2\ \mathrm{d}t}\\ i&=20+100\sin(100\pi t)\\ i^2&=400+4000\sin(100\pi t)+10000\sin^2(100\pi t)\\ &=400+4000\sin(100\pi t)+10000\times\frac 12(1-\cos(200\pi t))\\ \text{RMS}^2&=\frac1{0.02-0}\int_0^{0.02}(400+4000\sin(100\pi t)+10000\times\frac 12(1-\cos(200\pi t)))\ \mathrm{d}t\\ &=\frac1{0.02}\left[400t-\frac{4000\cos(100\pi t)}{100\pi}+5000t-\frac{5000\sin(200\pi t)}{200\pi}\right]_0^{0.02}\\ &=\frac1{0.02}\left[(8-12.732+100-0)-(0-12.732+0-0)\right] = \frac1{0.02}\times108\\ \text{RMS}&=\sqrt{\frac{108}{0.02}}=73.5\text{ A} \end{align*}\]

9.2 Volume and Surface of Revolution

Consider a curve \(y = f(x)\) between the ordinates \(x = a\) and \(x = b\). The volume of revolution is the volume of the equivalent solid body obtained by rotating the curve around the \(x\) or \(y\) axes.

9.2.1 Rotation about the \(x\)-axis

Figure 9.2: volume of revolution

If the shaded strip in Figure 9.2 is rotated about the \(x\)-axis then, provided \(\delta x\) is small, it will trace out a disc with thickness \(\delta x\) and volume \(\delta V = \pi y^2\delta x\), where \(y\) is the radius of the disc, see Figure 9.3.

Figure 9.3: volume of revolution

We may sum the volumes generated by all the strips between \(x = a\) and \(x = b\) to obtained the volume of the solid of revolution for the section of curve about the \(x\)-axis. Thus the total volume \(V\) is given by \[ V = \int_a^b\pi y^2\ \mathrm{d}x \]

9.2.2 Rotation about the \(y\)-axis

By a similar method it can be shown that for a rotation about the \(y\)-axis, the volume the solid of revolution will be \[ V = 2\pi\int_a^bxy\ \mathrm{d}x. \] This is for the volume of the solid created by rotating (about the \(y\)-axis) the same region as before (i.e. bounded by two vertical lines \(x=a\) and \(x=b\) on the sides, and the \(x\)-axis and the graph of a function \(y(x)\) from the bottom and the top) is rotated about the \(y\)-axis.

9.2.3 Surface area of revolution

The surface area of revolution is the surface area of the body generated by rotating a curve about the \(x\) or \(y\) axes. For a strip as shown in the figure above, we approximate the ‘arc’ (a part of the graph of the function) by a straight line. This yields a right triangle, from which we get an approximation for the length of arc to be \(\delta l=\sqrt{1+(\mathrm{d}y/\mathrm{d}x)^2}\cdot\delta x\).

If the strip is rotated about the \(x\)-axis, the surface area generated will be approximately \(\delta S = 2\pi y\delta l\). The surface area of the whole solid generated by rotating the curve around the \(x\)-axis between \(x = a\) and \(x = b\) is obtained by summing the contributions to the surface area from each of the elementary strips and we can eventually deduce that \[ S = 2\pi\int_a^b y\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\ \mathrm{d}x. \]

Example 9.3

The straight line \(y=mx\) between \(x=0\) and \(x=h\) is rotated about the \(x\)-axis thus generating a right circular cone. Calculate its volume in terms of \(h\) and \(R\) where \(R\) is the \(y\) value at \(x=h\). Find also the surface area of the resultant cone.

Figure 9.4: right-circular-cone

Solution. See Figure 9.4.

The volume of the solid of revolution for \(y = mx\) is given by \[ V = \pi\int_0^hy^2\ \mathrm{d}x = \pi\int_0^h (mx)^2\ \mathrm{d}x = m^2\pi\int_0^hx^2\ \mathrm{d}x = m^2\pi\left[\frac{x^3}3\right]_0^h = m^2\pi\frac{h^3}3. \] Now \(m = R/h\) and so \[ V = \frac13\frac{R^2}{h^2}h^3 = \frac13\pi R^2h. \] The surface area is given by \[\begin{align*} S&= 2\pi\int_0^hy\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\ \mathrm{d}x = 2\pi\int_0^h mx\sqrt{1+m^2}\ \mathrm{d}x\\ &= 2m\sqrt{1+m^2}\pi\left[\frac{x^2}2\right]_0^h = m\sqrt{1+m^2}\pi h^2 = \pi R h\sqrt{1+m^2}\\ &= \pi Rl. \end{align*}\] where \(l\) is the length of the inclined line.

Example 9.4 A curve is given parametrically as \(x=4t^2, y = 6t-t^2\). Determine the volume generated when the figure bounded by the curve, the \(x\)-axis and the ordinates corresponding to \(t=0\) and \(t=2\) is rotated about the \(x\)-axis.

Solution. \[ V = \pi\int_a^b y^2\ \mathrm{d}x. \] Since \(x=4t^2\) then \(\mathrm{d}x/\mathrm{d}t = 8t\) and so we can write \(\mathrm{d}x = 8t\ \mathrm{d}t\). Hence \[\begin{align*} V& = \pi\int_0^2(6t-t^2)^28t\ \mathrm{d}t = 8\pi\int_0^2(36t^3-12t^4+t^5)\ \mathrm{d}t\\ & = 8\pi\left[9t^4-\frac{12t^5}5+\frac{t^6}6\right]_0^2\\ &= 8\pi\times77.87 = 1957 \text{ units}^3. \end{align*}\]