Week 16 - Applications of Integration

1. Area between Curves

  • The area between the curves y=f(x)y=f(x) and y=g(x)y=g(x) and between x=ax=a and x=bx=b is A=abf(x)g(x)dxA=\int_a^b |f(x) - g(x)| dx
  • For example, f(x)=x2,g(x)=1,a=1,b=1f(x) = x^2, g(x) = 1, a = -1, b = 1
    • we can simply get the integral function 111x2 dx\int_{-1}^{1} 1 - x^2 \ dx
  • Some regions are best treated by regarding x as a function of y. If a region is bounded by curves with equations x=f(y)x=f(y), x=g(y)x=g(y), y=cy=c, and y=cy=c, where f and g are continuous and f(y)g(y)f(y) \ge g(y) for cydc \le y \le d, then its area is A=cdf(y)g(y)dyA=\int_c^d |f(y) - g(y)| dy
  • For example, f(x)=x,g(x)=2x1,a=0,b=1f(x)=\sqrt{x}, g(x) = \sqrt{2x-1}, a = 0, b = 1
    • If treat them as functions of x, we need to split the area to two parts.
    • So we rewrite the functions to f(y)=y2,g(y)=y2+12,c=0,d=1f(y) = y^2, g(y) = \frac{y^2 + 1}{2}, c = 0, d = 1

2. Volumes

2.1. Sphere's Volume

  • The cross-sectional area is A(x)=πy2=π(r2x2)A(x) = \pi y^2 = \pi(r^2 - x^2)
  • Using the definition of volume with a=ra = -r and b=rb = r, we have V=rrA(x)dx=rrπ(r2x2)dx=2π0r(r2x2) dx=2π[r2xx33]0r=2π(r3r33)=43πr3\begin{aligned} V &= \int_{-r}^{r}A(x) dx = \int_{-r}^{r} \pi(r^2 - x^2) dx \\ &= 2\pi \int_0^r(r^2 - x^2)\ dx \\ &= 2\pi \big[r^2x - \frac{x^3}{3}\big]_0^r = 2\pi(r^3-\frac{r^3}{3}) \\ &= \frac{4}{3}\pi r^3 \end{aligned}

2.2. Use Washer Method

  • The region R\mathscr{R} enclosed by the curves f(x)=x,g(x)=x2f(x) = x, g(x) = x^2 is rotated about the x-axis. Find the volume of the resulting solid.
  • The curves y=x and y=x2y = x \text{ and } y = x^2 intersect at the points (0, 0) and (1, 1). The region between them, the solid of rotation, and a cross-section perpendicular to the x-axis are shown above. A cross-section in the plane PxP_x has the shape of a washer (an annular ring) with inner radius x2x^2 and outer radius xx, so we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle: A(x)=πx2π(x2)2=π(x2x4)A(x) = \pi x^2 - \pi (x^2)^2 = \pi(x^2 - x^4)
  • Therefore we have V=01A(x)dx=01π(x2x4) dx=π[x33x55]01=2π15V = \int_0^1 A(x)dx = \int_0^1 \pi(x^2 - x^4) \ dx = \pi \big[\frac{x^3}{3} - \frac{x^5}{5}\big]_0^1 = \frac{2\pi}{15}

2.3. Use Shells Method

  • To face a situation like below. If we slice perpendicular to the y-axis, we get a washer. But to compute the inner radius and the outer radius of the washer, we’d have to solve the cubic equation f(x)f(x) for x in terms of y; that’s not easy.
  • For this situation, we use the method of cylindrical shells.
  • Instead of slicing the perpendicular to the x-axis:
  • Then, flatten as below:
    • Radius xx, circumference 2πx2 \pi x, height f(x)f(x), and thickness Δx\Delta x or dxdx:
  • For example: Find the volume of the solid obtained by rotating the region bounded by y=xx2y=x-x^2 and y=0y=0 about the line x=2x = 2.
    • The figure below shows the region and a cylindrical shell formed by rotation about the line x=2x = 2. It has radius 2x2 - x, circumference 2π(2x)2\pi (2-x), and height xx2x - x^2.
    • The volume of the given solid is: V=012π(2x)(xx2)dx=2π01(x33x2+2x)dx=2π[x44x3+x2]01=π2\begin{aligned} V &= \int_0^1 2 \pi (2 - x) (x - x^2) dx \\ &= 2 \pi \int_0^1 (x^3-3x^2 + 2x) dx \\ &= 2 \pi \big[\frac{x^4}{4} - x^3 + x^2\big]_0^1 = \frac{\pi}{2} \end{aligned}

2.4. Disks and Washers versus Cylindrical Shells

  • If the region more easily described by top and bottom boundary curves of the form y=f(x)y = f(x), or by left and right boundaries x=g(y)x = g(y), use Washers.
  • If we decide that one variable is easier to work with than the other, then this dictates which method to use.
  • Draw a sample rectangle in the region, corresponding to a cross-section of the solid. The thickness of the rectangle, either Δx\Delta x or Δy\Delta y, corresponds to the integration variable. If you imagine the rectangle revolving, it becomes either a disk (washer) or a shell.

3. Arc Length

  • Formula: If ff' is continuous on [a, b], then the length of the curve y=f(x),axby = f(x), a \le x \le b, is L=ab1+[f(x)]2dxL = \int_a^b \sqrt{1+[f'(x)]^2}dx
  • Proves:
    • Suppose that a curve C is defined by the equation y=f(x)y = f(x), where ff is continuous and axba \le x \le b. We obtain a polygonal approximation to C by dividing the interval [a, b] into n subintervals with endpoints x0,x1,,xnx_0, x_1, \ldots, x_n and equal width Δx\Delta x. If yi=f(xi)y_i = f(x_i), then the point Pi(xi,yi)P_i(x_i, y_i) lies on C and the polygon with vertices P0,P1,,PnP_0, P_1, \ldots, P_n, is an approximation of C.
    • We define the length L of the curve C as thee limit of the lengths Pi1Pi|P_{i-1}P_i|: L=limni=1nPi1PiL = \lim_{n \to \infty} \sum_{i=1}^n |P_{i-1}P_i|
    • If we let Δyi=yiyi1\Delta y_i = y_i - y_{i-1}, then Pi1Pi=(xixi1)2+(yiyi1)2=(Δxi)2+(Δyi)2=Δxi1+(ΔyiΔxi)2=1+(f(x))2dx\begin{aligned} |P_{i-1}P_i| &= \sqrt{(x_i-x_{i-1})^2 + (y_i-y_{i-1})^2} = \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2} \\ &= \Delta x_i \sqrt{1 + (\frac{\Delta y_i}{\Delta x_i})^2} \\ &= \sqrt{1 + (f'(x))^2}dx \end{aligned}
    • So L=ab1+[f(x)]2dxL = \int_a^b\sqrt{1 + [f'(x)]^2}dx

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