curve(2*x-x^2, col="blue", lwd=2, ylab = "f(x)" )
curve(x^4, add=TRUE, col="red", lwd=2)
legend(x="right",legend = c("f(x)","g(x)"), col=c("blue","red"),lty = 1, lwd=2,cex=0.6 )
Chapter 6 Applications of the Integral
6.1 The Area of a Plane Region
We use the definite integral to calculate areas of regions of more and more complicated shapes.
A Region above and below the \(x\)-Axis
To find a region above the \(x\)-axis, we use the formula of area \(A(R)\) of a region \(R\) (the region under \(y=f(x)\) between \(x=a\) and \(x=b\)) \[A(R)=\int_a^bf(x)dx\]
Example 1, page 335.
Note that when the graph of \(f(x)\) is below the \(x\)-axis, \(\int_a^bf(x)dx\) is a negative number and therefore cannot be an area. However, it is just the negative of the area of the region bounded by \(y=f(x)\), \(x=a\), \(x=b\) and \(y=0\).
Example 2 & 3, page 335-336.
A Helpful Way of Thinking
For simple regions of the type considered above, it is quite easy to write down the correct integral. When we consider more complicated regions, the task of selecting the right integral is more difficult. Here is a way of thinking that can be very helpful. It goes back to the definition of area and of the definite integral.
Example 4, page 336.
Step 3: The area of a typical piece is approximated by \[\Delta A_i=f(x_i)\Delta x_i=(1+\sqrt {x_i})\Delta x_i\]
Step 4: Add up. \(A=\sum \Delta A_i\)
Step 5: Take the limit, i.e., the definite integral on the interval.
A Region Between Two Curves
We still use the slice, approximate, integrate method to find its area. Be sure to note that \(f(x)-g(x)\) gives the correct height for the thin slice, even when the graph of \(g\) goes below the \(x\)-axis.
N.B. Be sure to specify which one is \(f(x)\) and which is \(g(x)\).
Example 5, page 337. Find the area of the region between the curves \(y=x^4\) and \(y=2x-x^2\).
We start by finding where the two curves intersect. \((0, 0)\) and \((1, 1)\). So, we evaluate the definite integral on \([0,1]\)\[ \int_0^1 (2x-x^2-x^4)dx \]
# find the root of polynomial equation: 2x- x^2-x^4=0
# vector of polynomial coefficients in increasing order
z <- c(0, 2,-1, 0, -1)
polyroot(z) # two real roots: x = 0 or 1[1] 0.0+0.000000i 1.0+0.000000i -0.5+1.322876i -0.5-1.322876i
Vertical Slicing and Horizontal Slicing
Example 6, page 337. Find the area of the region between the parabola \(y^2=4x\) and the line \(4x-3y=4\).
Now imagine slicing this region vertically. We face a problem, because the lower boundary consists of two different curves. Slices at the extreme left extend from the lower branch of the parabola to its upper branch. To solve the problem with vertical slices requires that we first split our region into two parts, set up an integral for each part, and then evaluate both integrals.
Vertical Slicing
\[\begin{aligned}&\int_0^{\frac{1}{4}}\left[2\sqrt x-(-2\sqrt x)\right]dx + \int_{\frac{1}{4}}^4\left[ 2\sqrt x-(\frac{4}{3}x-\frac{4}{3})\right] dx \\ =&\frac{8}{3}x^{\frac{3}{2}}\Bigg |_0^{\frac{1}{4}} +\frac{4}{3} x^{\frac{3}{2}}\Bigg |_\frac{1}{4}^{4}-\frac{2}{3} x^{2}\Bigg |_\frac{1}{4}^{4}+\frac{4}{3} x\Bigg |_\frac{1}{4}^{4}\\=& \frac{8}{3}\times (\frac{1}{2})^3+\frac{4}{3}\times(8-\frac{1}{8})-\frac{2}{3}\times(16-\frac{1}{16})+\frac{4}{3}\times(4-\frac{1}{4})\\=&\frac{125}{24}\end{aligned}\]
Horizontal Slicing
N.B. There are two items to note for the horizontal slicing: (1) The integrand resulting from a horizontal slicing involves \(y\), not \(x\); and (2) to get the integrand, solve both equations for \(x\) and subtract the smaller \(x\)-value from the larger.
Distance and Displacement – Revisit
N.B. Displacement measures the change in position. Displacement is the directed distance.
6.2 Volumes of Solids: Slabs, Disks, Washers
We start with simple solids called right cylinders, four of which are shown in Figure 1. In each case, the solid is generated by moving a plane region (the base) through a distance \(h\) in a direction perpendicular to that region. And in each case, the volume of the solid is defined to be the area \(A\) of the base times the height \(h\); that is, \[V=A\cdot h\]
Next, we consider a solid with the base (or cross-section) of non-constant area \(A(x)\), which varies with \(x\). We slice the solid into thin slabs (in Figure 3), each slab is thin enough to be regarded as a right cylinder. Therefore, the volume of a slab is approximately \[\Delta V_i\approx A(\bar x_i)\Delta x_i\] where \(\Delta x_i\) is the length of the partition interval, \(\bar x_i\) is the sample point of the interval \([x_{i-1}, x_i]\).
Rather than routinely applying the boxed formula to obtain volumes, we suggest that in each problem you go through the process that led to it. Just as for areas, we call this process slice, approximate, integrate. It is illustrated in the examples that follow.
Solids of Revolution: Method of Disks
When a plane region, lying entirely on one side of a fixed line in its plane, is revolved about that line, it generates a solid of revolution. The fixed line is called the axis of the solid of revolution.
We could represent the volume of a solid of revolution as a definite integral.
Example 1, page 343. Find the volume of the solid of revolution obtained by revolving the plane region \(R\) bounded by \(y=\sqrt x\), the \(x\)-axis, and the line \(x=4\) about the \(x\)-axis.
Step 1 Slice. See Figure 7.
Step 2 Approximate: The volume of solid is approximated by the Riemann sum of the volume of the stacking slabs \[V\approx\sum_{i=1}^nA(\bar x_i)\Delta x_i=\sum_{i=1}^n\pi (\sqrt x)^2 \Delta x\]
Step 3: Integrate. \[V=\int_0^4 \pi (\sqrt x)^2dx\]
Example 8, page 343. Find the volume of the solid generated by revolving the region bounded by the curve \(y=x^3\), the y-axis, and the line \(y=3\) about the y-axis (Figure 8).
N.B. The solid is generated by revolving the plane about the \(y\)-axis. Here we slice horizontally, that is, the thin slabs stacking vertically, which makes \(y\) the choice for the integration variable.
Method of Washers
Sometimes, slicing a solid of revolution results in disks with holes in the middle.We call them washers.
Example 3, page 344.
Example 4, page 344.
Specify \(r_1\) and \(r_2\) by the graph. Note that the revolving axis is \(x=-1\).
\[\begin{aligned} V&=\int_{-2}^2\pi\left[\left(1+\sqrt{4-y^2}\right)^2-1^2\right]dy\\&=2\pi\int_{0}^2\left[2\sqrt{4-y^2}+4-y^2\right]dy\\&=2\pi\left[2\underbrace{\int_{0}^2 \sqrt{4-y^2}dy}_{\text{area of a quarter circle with radius }2\text{ is }\pi}+\int_{0}^2(4-y^2)dy\right]\\&=2\pi\left[2\pi+\left(4y-\frac{1}{3}y^3\right)\Bigg|_0^2\right]\\&= 2\pi\left(2\pi+\frac{16}{3}\right)\end{aligned}\]
Other Solids with Known Cross Sections
This slice, approximate, integrate procedure can be used to find the volume of a solid whose area of the cross section can be determined.
Example 5, page 345.
Note that the cross sections perpendicular of this solid are squares.
Example 6, page 346.
Note that the cross sections are equilateral triangles. Recall the area of an equilateral triangle of side \(a\) is \(\frac{\sqrt 3}{4}a^2\).
6.3 Volumes of Solids of Revolution: Shells
There is another method for finding the volume of a solid of revolution: the method of cylindrical shells. For many problems, it is easier to apply than the methods of disks or washers.
The Method of Shells
Again, the strategy is slice, approximate, integrate.
Example 1, page 349.
Step 1, slice. Like figure 3.
Step 2, approximate. Use the formula, the volume of the shell is approximated by the sum of the slices \[V\approx \sum_{i=1}^n 2\pi x_i f(x_i)\Delta x_i\]
Step 3, integrate. \[V=\int_1^4 2\pi x \frac{1}{\sqrt x}dx\]
Disk Method vs. Shell Method
Example 2, page 350.
Disk Method: The revolving axis is \(x\)-axis. The integrate variable is \(x\).
Shell Method: The integrate variable is \(y\).
\[V=\int_a^b 2\pi \left(\text{ shell radius }\right)\left( \text{ shell height }\right)dy\\=\int_0^r 2\pi y\left(h-\frac{h}{r}y\right)dy\]
Example 3, page 350. Find the volume of the solid generated by revolving the region in the first quadrant that is above the parabola \(y=x^2\) and below the parabola \(y=2-x^2\) about the \(y\)-axis.
The disk method are not the best choice (because the right boundary consists of parts of two curves, making it necessary to use two integrals).
Putting It All Together
In the next example, we are going to imagine revolving the region \(R\) of Figure 7 about various axes. Our job is to set up and evaluate an integral for the volume of the resulting solid, and we are going to do it by looking at a plane figure.
N.B. In all four cases the limits of integration are the same; it is the original plane region that determines these limits.
6.4 Length of a Plane Curve
- Plane Curve
-
A plane curve is determined by a pair of parametric equations \(x=f(t)\), \(y=g(t)\), \(a\leq t\leq b\), where we assume that \(f\) and \(g\) are continuous on the given interval \([a,b]\). As \(t\) increases from \(a\) to \(b\), the point \((x,y)\) traces out a curve in the plane.
For example, tracing the the points of the parabola in parametric form \(x=2t+1, y=t^2-1, 0\leq t\leq 3\), is visualized as follows
The direction indicated by an arrow on the curve in Figure 5, is called the orientation of the curve.
t <- seq(0,3, by=.2)
index.t <- seq(1, length(t), 5) # indexing every 5 values
x <- 2*t+1
y <- t^2-1
# Table
cbind("t"=1:length(index.t)-1,"x"=x[index.t], "y"=y[index.t]) t x y
[1,] 0 1 -1
[2,] 1 3 0
[3,] 2 5 3
[4,] 3 7 8# Sketch the Curve
plot(x,y,type="l")
points(x[index.t], y[index.t], col="blue", pch=16)
For example, the parametric equations \(x=t-\sin t, y=1-\cos t, 0\leq t\leq 4\pi\) are not smooth curve, because \(dx/dt= 1-\cos t\) and \(dy/dt=\sin t\) are simultaneously \(0\) when \(t=2\pi\).
Arc Length
How could we find the length of curve given by \(x=f(t)\), \(y=g(t)\), \(a\leq t\leq b\)?
Example 3, page 357. Polar Coordinate.
Example 4, page 357. Find the length of the line segment from \(A(0, 1)\) to \(B(5, 13)\).
By the slope-point formula, we obtain the line \(y=\frac{12}{5}x+1\). Since \(\frac{dy}{dx}=\frac{12}{5}\) and \(0\leq x\leq 5\), it is obvious we use the second formula to find the arc length.
Comment: For most arc length problems it is easy to set up the definite integral, but it is often difficult or impossible to evaluate these integrals. Numerical technique is used to solve such problems, but that is beyond the scope of this course. So, you can just skip Example 6 while reading the textbook.
Differential of Arc Length
Based on the First Fundamental Theorem of Calculus, it is natural to obtain the differential of arc length, given the formula of the arc length.
Area of a Surface of Revolution (Optional)
If a smooth plane curve is revolved about an axis in its plane, it generates a surface of revolution as illustrated in Figure 13. Our aim is to determine the area of such a surface.
The surface area is given by
