In this chapter, we present two applications of the definite integral: finding the area between curves in the plane and finding the volume of the 3D objected obtained by rotating about some given axis the area between curves.
Areas between curves
Consider finding the area between two given curves, say \(y=f(x)\) and \(y=g(x)\), over an interval \(a \leq x \leq b\):
Area between two curves
We will first estimate the area and then invoke a limiting process to argue that in the limit our approximations converge to the true area. To start, subdivide the interval \([a,b]\) into \(n\) equal subintervals of length \(\Delta x = \frac{b-a}{n}\). The subdivision process produces the intervals \([x_0,x_1]\), \([x_1,x_2]\), \(\ldots\), \([x_{n-1}, x_n]\). Consider one of these subinterval, say \([x_{k-1}, x_k]\), and approximate on this subinterval the area between the curves with a rectangle of area
\[
R_k = \text{Area of \(k\)th rectangle} = \underbrace{(f(c_k)-g(c_k))}_{\text{height}}\underbrace{\Delta x}_{\text{width}}
\]
where \(c_k\) is a point chosen inside the interval \([x_{k-1}, x_k]\). The total area between the curves can be approximated by adding the areas of all the rectangles (see Figure 1.2):
\[
\text{Exact Area} \approx \text{Area of \(n\) rectangles} = \sum_{k=1}^n (f(c_k)-g(c_k)) \Delta x.
\]
The expression below is called a Riemann sum:
\[
\sum_{k=1}^n (f(c_k)-g(c_k)) \Delta x = (f(c_1)-g(c_1))\Delta x + \cdots + (f(c_n)-g(c_n))\Delta x.
\]
Using rectangles to approximate areas between curves; in this case we subdivide the interval \([a,b]\) into \(n=8\) equal subintervals and over each subinterval we approximate the area between the curves with the area of a rectangle.
If now we increase the value of \(n\) and repeat the process we obtain a new approximation which is closer to the true area between the curves. Hence, to compute exact area we take the limit of the Riemann sum expression as \(n\rightarrow\infty\). As you may recall from Calculus I, the limit of a Riemann sum is the definition of the integral:
\[
\text{Exact Area} = \lim_{n\rightarrow\infty} \sum_{k=1}^n (f(c_k)-g(c_k)) \Delta x = \int_a^b (f(x) - g(x))\, dx.
\]
It is important to emphasize that on the interval of integration \([a,b]\), we need
\[
f(x) \geq g(x).
\]
In summary, the exact area \(A\) can be computed as the integral
\[
A = \int_a^b [\underbrace{f(x)}_{\text{top}} - \underbrace{g(x)}_{\text{bottom}} ] \, dx.
\]
Find the area of the region bounded by the curves \(y=e^x\), \(y=x\), \(x=0\), and \(x=1\).
The curves and the region are shown below
Curves for Example 1.1.1
The area is then
\begin{align*}
\text{A} &= \int_0^1(e^x - x)\, dx\\
&= (e^x - x^2)/2) \Big|_0^1 \\
&= e - \tfrac{3}{2}
\end{align*}
Find the area of the region bounded by the curves \(y=x^2\) and \(y=2x-x^2\).
The curves and region is shown below. The curves intersect when \(x^2 = 2x - x^2\) or when \(x=0\) and \(x=1\). The area is then
\begin{align*}
\text{A} &= \int_0^1 (2x-x^2 - x^2) = \frac{1}{3}
\end{align*}
Curves for Example 1.1.2
Find the area of the region bounded by the curves \(y=x^3-2x\) and \(y=-x^2\).
The curves and region is shown below. The points of intersection are found by setting \(x^3-2x = -x^2\) and this gives
\[
x (x+2)(x-1) = 0
\]
The \(x\)-values where the curves intersect are therefore \(x=0\), \(x=-2\), and \(x=1\). We must do two integrals to compute the area.
\begin{align*}
\text{A} &= \int_{-2}^0 [(x^3-2x)-(-x^2)]\, dx + \int_0^1 [-x^2-(x^3-2x)]\, dx\\
&= \frac{8}{3} + \frac{5}{12} = \frac{37}{12}
\end{align*}
Curves for Example 1.1.3
Find the area of the region bounded by the curves \(y=x^2\), \(y=2-x\), and \(y=0\).
The curves and region are shown below. If we integrate with respect to \(x\), we will need to do two integrals. On the other hand, the region is bounded on the right by the curve \(y=2-x\) and on the left by the curve \(y=x^2\). The curves intersect at \(x=1\) and \(y=1\). Integrating with respect to \(y\):
\begin{align*}
\text{A} &= \int_0^1 [\text{Right}(y) - \text{Left}(y)]\, dy \\
&= \int_0^1 [2-y - \sqrt{y} ] \, dy\\
&= [2y - \tfrac{1}{2}y^2 - \tfrac{2}{3}y^{3/2}]\Big|_0^1 \\
&= 2 - \tfrac{1}{2} - \tfrac{2}{3} \\
&= \tfrac{5}{6}
\end{align*}
Curves for Example 1.1.4
Find the area of the region bounded by the curves \(2y=5\sqrt{x}\), \(y=4\), and \(2y+x=6\).
The region and curves is shown below. To find where the curves intersect, set \(\frac{4}{25} y^2 = 6-2y\) which yields the quadratic
\[
2y^2 + 25y - 75 = 0
\]
whose roots are \(y=\tfrac{5}{2}\) and \(y=-15\). Only interested in \(y=\tfrac{5}{2}\) intercept. Integrating with respect to \(y\) yields
\begin{align*}
\text{A} &= \int_{5/2}^4[\tfrac{4}{25}y^2 - (6-2y)]\, dy\\
&= \tfrac{4}{75}y^3 - 6y + y^2 \Big|_{5/2}^4 \\
&= \frac{333}{100}
\end{align*}
If instead integrating with respect to \(x\) need two integrals:
\begin{align*}
\text{A} &= \int_{-4}^1 [4-(-x/2 + 3)] \, dx + \int_0^{\tfrac{64}{25}} [4-\tfrac{5}{2}\sqrt{x}]\,dx\\
&= 9/4 + 27/25 = 333/100
\end{align*}
Curves for Example 1.1.5
Find the area of the region bounded by \(y=\sin(x)\) and \(y=\cos(x)\) over the interval \([0,\pi]\).
The region and curves is shown below. On the interval \([0,\pi]\), sine and cosine intersect at \(x=\pi/4\). Need two integrals:
\begin{align*}
\text{Area} &= \int_0^{\pi/4} [\cos(x)-\sin(x)]\,dx + \int_{\pi/4}^\pi(\sin(x)-\cos(x))\,dx\\
&= (\sin(x)+\cos(x))\Big|_0^{\pi/4} + (-\cos(x)-\sin(x))\Big|_{\pi/4}^\pi\\
&= \sqrt{2}-1 + 1+\sqrt{2} = 2\sqrt{2}
\end{align*}
Curves for Example 1.1.6
Find the area of the region bounded by \(y=\frac{1}{x}\), \(y=4x-3\), and \(y=5\).
The region and curves is shown below. Best handled with integration with respect to \(y\):
\begin{align*}
\text{A} &= \int_1^5 [(y+3)/4 - 1/y]\,dx\\
&= \tfrac{1}{8}y^2 + \tfrac{3}{4}y - \ln|y| \Big|_1^5 \\
&= \tfrac{25}{8} +\tfrac{15}{4} - \ln(5) - \tfrac{1}{8} - \tfrac{3}{4}\\
&= 6 -\ln(5)
\end{align*}
Curves for Example 1.1.7
Volumes
Before beginning this subject, see the Maple file suface-revolution.mw found in the directory
Teaching-Resources/Calculus-Resources/Maple
In this section, we will compute volumes of solids using the integral. First recall that the volume of a cylinder is \(V = \pi r^2 h\), see Figure 1.10.
Volume of a cylinder of radius \(r\) and height \(h\) is \(V = \pi r^2 h\)
Consider a function \(f\) on the interval \([a,b]\) and rotate around the \(x\)-axis the region under the graph of \(y=f(x)\), between \(x=a\) and \(x=b\), as shown below:
We can estimate the volume of the solid using discs as follows. We subdivide the interval \([a,b]\) into \(n\) equal subintervals of length \(\Delta x = \frac{b-a}{n}\). This produces grid points along the interval \([a,b]\):
\[
a = x_0 \lt x_1 \lt x_2 \lt \cdots \lt x_n = b
\]
On the \(k\)th generic interval \([x_{k-1}, x_k]\), we choose the mid-point \(c_k = \frac{x_k + x_{k-1}}{2}\) and consider the disc of radius \(f(c_k)\) and width \(\Delta x\), whose volume is therefore
\[
\pi r^2 h = \pi f(c_k)^2 \Delta x
\]
We repeat this process at every interval and obtain \(n\) discs whose volumes are added to obtain an approximation to the true volume of the region, such a sum is called a Riemann sum:
\[
V \approx \sum_{k=1}^n \pi f(c_k)^2 \Delta x
\]
As \(n\) increases and we use more discs with smaller width, we obtain a sequence of values that converge to the true volume of the solid:
\[
V = \lim_{n\rightarrow\infty} \sum_{k=1}^n \pi f(c_k)^2 \Delta x
\]
The limit of a Riemann sum is an integral and thus the volume of the solid is
\[
V = \int_a^b \pi f(x)^2 \, dx.
\]
Show that the volume of a sphere of radius \(R\) is \(V=\frac{4}{3}\pi R^3\).
A sphere can be obtained by rotating about the \(x\)-axis a semi-circle of radius \(R\). The function whose graph is the upper semi-circle is \(f(x)=\sqrt{R^2-x^2}\). Therefore, the volume of the sphere is
\begin{align*}
V &= \int_{-R}^R \pi (\sqrt{R^2-x^2})^2 \, dx\\
&= \int_{-R}^R \pi (R^2-x^2) \, dx\\
&= (R^2 x - \tfrac{1}{3}x^3)\Big|_{-R}^R\\
&= \frac{4}{3}\pi R^3.
\end{align*}
Find the volume of the solid obtained by rotating about the \(x\)-axis the region under the curve \(y=\sqrt{x}\) from \(x=0\) to \(x=1\).
Below is the solid and a generic disc. The volume is
\[
V = \int_0^1 \pi (\sqrt{x})^2\, dx = \pi \int_0^1 x\, dx = \frac{\pi}{2}
\]
Find the volume of the solid obtained by rotating about the \(y\)-axis the region bounded by \(y=x^3\), \(y=8\), and \(x=0\).
In this case, we need to use horizontal discs. At a value of \(y\) between \([0,8]\), a typical disc has radius \(r = y^{1/3}\). The volume is therefore
\begin{align*}
V &= \int_0^8 \pi (y^{1/3})^2 \, dy\\
& = \pi \frac{y^{5/8}}{5/3}\Big|_0^8 \\
&= \frac{3\pi}{5} (8^{1/3})^5\\
&= \frac{96\pi}{5}
\end{align*}
Find the volume of the solid obtained by rotating about the \(x\)-axis the region bounded by the curves \(y=\sqrt{x}\) and \(y=x^2\).
The solid is shown below. In this case, instead of discs we must use washers. The inner radius is \(r_{\text{in}} = x^2\) and the outer radius is \(r_{\text{out}} = \sqrt{x}\). The volume of a typical washer is having width \(\Delta x\) is therefore
\[
V = \pi r_{\text{out}}^2 \Delta x - \pi r_{\text{in}}^2 \Delta x = \pi [r_{\text{out}}^2 - r_{\text{in}}^2] \Delta x
\]
Therefore, the volume of the solid is
\begin{align*}
V &= \int_0^1 \pi [(\sqrt{x})^2 - (x^2)^2] \, dx\\
&= \int_0^1 \pi [x - x^4]\, dx\\
&= \pi (\tfrac{1}{2}x^2 - \tfrac{1}{5}x^5)\Big|_0^1\\
&= \tfrac{3\pi}{5}
\end{align*}
Find the volume of the solid obtained by rotating about the \(y\)-axis the region bounded by the curves \(y=x^3\), \(x=0\), \(x=2\), and \(y=0\).
At a \(y\) value between \([0,8]\), a typical washer has inner radius \(r_{\text{in}} = y^{1/3}\) and an outer radius \(r_{\text{out}} = 2\). The volume is therefore
\begin{align*}
V &= \int_0^8 \pi [2^2 - (y^{1/3})^2] \, dy\\
&= \int_0^8 \pi (4 - y^{2/3})\, dy\\
&= \pi (4y - \tfrac{3}{5}y^{5/3})\Big|_0^8\\
&= \frac{64\pi}{5}
\end{align*}
Find the volume of the solid obtained by rotating about the line \(y=2\) the region bounded by the curves \(y=e^{-x}\), \(y=1\), and \(x=2\).
The inner radius of a generic washer is \(r_{\text{in}} = 1\) and outer radius is \(r_{\text{out}} = 2-e^{-x}\). The volume is therefore
\begin{align*}
V &= \int_0^2 \pi [r_{\text{out}}^2 - r_{\text{in}}^2]\, dx\\
&= \int_0^2 \pi [ (2-e^{-x})^2 - 1^2 ] \, dx\\
&= \int_0^2 \pi (3-4e^{-x} + e^{-2x})\,dx\\
&= \pi (3x+4e^{-x} - \tfrac{1}{2}e^{-2x})\Big|_0^2\\
&= \pi (\tfrac{5}{2} + 4e^{-2} - \tfrac{1}{2}e^{-4})
\end{align*}
Find the volume of the solid obtained by rotating about the line \(y=-5\) the region bounded by the curves \(y=\frac{1}{x^3}\), \(y=0\), \(x=1\) and \(x=2\).
The inner radius of a generic washer is \(r_{\text{in}} = 5\) and outer radius is \(r_{\text{out}} = 5 + \frac{1}{x^3}\). The volume is therefore
\begin{align*}
V &= \int_1^2 \pi \left[ (5+\frac{1}{x^3})^2 - 5^2\right] \, dx\\
&= \pi \int_1^2 \left(\frac{10}{x^3} + \frac{1}{x^6}\right)\, dx\\
&= \pi \left( \frac{-5}{x^2} - \frac{1}{5x^5}\right)\Big|_1^2\, dx\\
&= \frac{631}{160}\pi
\end{align*}
Find the volume of the solid obtained by rotating about the line \(x=4\) the region bounded by the curves \(x=y^2\) and \(x=2y\).
The inner radius of a generic washer is \(r_{\text{in}} = 4-2y\) and outer radius is \(r_{\text{out}} = 4-y^2\). The volume is therefore
\begin{align*}
V &= \int_0^2 \pi \left[ (4-y^2)^2-(4-2y)^2\right] \, dx\\
&= \pi \int_0^2 \left(y^4-12y^2+16y\right)\, dx\\
&= \frac{32}{5}\pi
\end{align*}
Find the volume of the solid obtained by rotating about the line \(x=3\) the region bounded by the curves \(y=x^2\) and \(y=3x\).
The inner radius of a generic washer is \(r_{\text{in}} = 3-\sqrt{y}\) and outer radius is \(r_{\text{out}} = 3-y/3\). The volume is therefore
\begin{align*}
V &= \int_0^9 \pi \left[ (3-y/3)^2-(3-\sqrt{y})^2\right] \, dx\\
&= \pi \int_0^9 \left(\frac{y^2}{9}-3y+6\sqrt{y}\right)\, dx\\
&= \frac{27}{2}\pi
\end{align*}
Arc Length
Before we begin, we recall the Mean Value Theorem.
If \(f\) is continuous on the interval \([a,b]\) and differentiable on \((a,b)\) then there is some number \(c\) in \((a,b)\) such that
\[
\frac{f(b)-f(a)}{b-a} = f'(c).
\]
Consider finding the length of a curve that is the graph of a function \(f\) over the interval \([a,b]\). Finding the length of a line segment is straightforward and thus we will first use line segments to approximate the graph of \(f\) and then use the sum of the lengths of the line segments to approximate the true length of the curve. This process will result in a Riemann sum which will then lead us to an integral for computing the length of the curve. Hence, to begin first subdivide the interval \([a,b]\) into \(n\) equal subintervals of length \(\Delta x = \frac{b-a}{n}\). On each subinterval, use a line segment to approximate \(f\) as shown below:
On each subinterval, the length of the line segment can be found by using the Pythagorean theorem:
\[
\ell_i = \sqrt{(\Delta x)^2 + (\Delta y)^2}
\]
where \(\Delta y = f(x_i) - f(x_{i-1})\).
Applying the Mean Value theorem to \(f\) on the interval \([x_{i-1},x_i]\), there is some point \(c_i\) such that
\[
\Delta y = f(x_i) - f(x_{i-1}) = f'(c_i) \Delta x.
\]
Therefore, the length of the line segment on the interval \([x_{i-1},x_i]\) is
\[
\ell_i = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{1+(f'(c_i))^2} \Delta x.
\]
Adding up the lengths of each line segment gives an approximation of the arc length
\[
L \approx \sum_{i=1}^n \sqrt{1+(f'(c_i))^2} \Delta x
\]
and taking the limit of the Riemann sum produces the exact value of \(L\):
\[
L = \lim_{n\rightarrow\infty} \sum_{i=1}^n \sqrt{1+(f'(c_i))^2} \Delta x.
\]
By definition, the integral is the limit of a Riemann sum and therefore
\[
L = \int_a^b \sqrt{1+[f'(x)]^2} \, dx.
\]
Let \(f(x) = 2x^{3/2}\). Find the arc length of the graph of \(f\) over the interval \([0,1]\).
The graph of the function is shown below:
Here \(f'(x) = 3x^{1/2}\) and the arc length is then
\begin{align*}
L &= \int_0^1 \sqrt{1+(3x^{1/2})^2}\, dx \\
&= \int_0^1 \sqrt{1+9x}\, dx u=1+9x\\
&= \frac{1}{9}\int_1^{10} \sqrt{u}\, du \\
&= \frac{1}{9} \left(\frac{u^{3/2}}{3/2} \right)\Big|_1^{10}\\
&= \frac{2}{27}(10^{3/2}-1)
\end{align*}
Find the arc length of the graph of the function \(f(x)=\ln(\sec(x))\) on the interval \([0,\pi/4]\).
The graph of the function is shown below:
Here \(f'(x) = \tan(x)\) and the arc length is therefore
\begin{align*}
L &= \int_0^{\pi/4} \sqrt{1+\tan^2(x)} \, dx\\
&= \int_0^{\pi/4} \sqrt{\sec^2(x)}\, dx\\
&= \int_0^{\pi/4} \sec(x)\,dx\\
&= \ln|\sec(x)+\tan(x)|\Big|_{0}^{\pi/4}\\
&= \ln|2/\sqrt{2}+1| - \ln|1+0|\\
&= \ln|2/\sqrt{2}+1|
\end{align*}
Average Value of a Function
The average value of a given set of numbers \(y_1, y_2, \ldots, y_n\) is
\[
y_{\text{avg}} = \frac{y_1+y_2+\cdots+y_n}{n} = \frac{1}{n} \sum_{i=1}^n y_i.
\]
What about the average value of a function \(y=f(x)\) over an interval \([a,b]\)? To get an approximation to the average value of \(f\), we could subdivide the interval \([a,b]\) into \(n\) equally wide subintervals \([x_0,x_1]\), \([x_1, x_2]\), \(\ldots\), \([x_{n-1},x_n]\), evaluate \(f\) at the end-points of each subinterval, \(y_i = f(x_i)\), and find the average of the resulting numbers:
\[
f_{\text{avg}} \approx \frac{1}{n}\sum_{i=1}^n f(x_i).
\]
The length of each subinterval is \(\Delta x = \frac{b-a}{n}\) and thus \(\frac{1}{n} = \frac{1}{b-a} \Delta x\). Thus,
\[
f_{\text{avg}} \approx \frac{1}{n}\sum_{i=1}^n f(x_i) = \frac{1}{b-a} \sum_{i=1}^n f(x_i) \Delta x.
\]
As \(n\rightarrow\infty\) we obtain the average value of \(f\), which yields the formula
\[
f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x) \, dx.
\]
On April 7, 2003 in Geneseo, the temperature (in degrees Fahrenheit) \(t\) hours after 9 am was modeled by the function \(T(t) = 50+14\sin(\pi t / 12)\). Find the average temperature during the period from 9 am to 9 pm on April 7, 2003.
The time \(t=0\) corresponds to 9 am, and thus 9 pm corresponds to \(t=12\). The average temperature was
\begin{align*}
T_{\text{avg}} &= \frac{1}{12}\int_0^{12} (50 + 14\sin(\pi t/12))\,dt \\
&= \frac{1}{12}\left[ 50t - \frac{12\cdot 14}{\pi} \cos(\pi t/12)\right]\Big|_0^{12}\\
&= 50 + \frac{28}{\pi} \\
&\approx 59\; F
\end{align*}
