• Area of a Region Between Two Curves
  • Volume
  • Arc Length

# Area of a Region Between Two Curves

The area A of the region bounded by the curves y=f(x),y=g(x)y =f(x), y= g(x) and the lines x=a,x=bx=a, x=b,
where ff and gg are continuous ,is A=abf(x)g(x)dxA = \int^b_a |f(x) - g(x)|dx

# Volume

# The Disk Method

  • V=abA(x)dxV = \int^b_a A(x)dx(about x-axis)
  • V=cdA(y)dyV = \int^d_c A(y)dy(about y-axis)

# The Shell Method

  • V=ab2πxf(x)dxV = \int^b_a 2 \pi xf(x) dx (about y-axis)
  • V=cd2πyf(y)dyV = \int^d_c 2 \pi yf(y) dy (about x-axis)

Find the volume by rotating about the y-axis the region bounded by y=2x2x3y = 2x^2 - x^3 and y=0y = 0.

V=02(2πx)(2x2x3)dx=2π022x3x4dx=2π12x415x502=2π(8325)=165πV = \int^2_0 (2 \pi x)(2x^2-x^3)dx = 2\pi \int^2_0 2x^3-x^4 dx = 2\pi \frac{1}{2}x^4 - \frac{1}{5}x^5|^2_0 = 2\pi (8-\frac{32}{5}) = \frac{16}{5} \pi

# Arc Length

PiPi+1=(xi+1xi)2+(f(xi+1)f(xi))2=(Δx)2+(Δy)2\overline{P_iP_{i+1}} = \sqrt{(x_{i+1}-x_i)^2 + (f(x_{i+1}) - f(x_i))^2} = \sqrt{(\Delta x)^2 + (\Delta y)^2}
i=1nPiPi+1=i=1n(Δx)2+(Δy)2=i=1n((Δx)2(Δx)2)+(Δy)2(Δx)2Δx=xixn1+(f(x))2dx\Rightarrow \sum^n_{i=1} \overline{P_iP_{i+1}} = \sum^n_{i=1} \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sum^n_{i=1} \sqrt{(\frac{(\Delta x)^2}{(\Delta x)^2}) + \frac{(\Delta y)^2}{(\Delta x)^2}} \Delta x= \int^{x_n}_{x_i} \sqrt{1 + (f'(x))^2 } dx

Assume that ff' is continuous on [a,b][a,b] and gg' is continuous on [c,d][c,d].

  1. If the arc is described by y=f(x)y=f(x),axba \le x \le b then L=ab1+(dydx)2dx=ab1+(f(x))2dxL = \int^b_a \sqrt{1+(\frac{dy}{dx})^2}dx = \int^b_a \sqrt{1+(f'(x))^2}dx
  2. If the arc is described by x=g(y)x=g(y),cydc \le y \le d then L=cd1+(dydx)2dy=cd1+(g(y))2dyL = \int^d_c \sqrt{1+(\frac{dy}{dx})^2}dy = \int^d_c \sqrt{1+(g'(y))^2}dy
  1. y2=x3y^2=x^3,Find the length of arc from (1,1)(1,1) to (4,8)(4,8).
  2. y2=xy^2 = x,Find the length of arc from (0,0)(0,0) to (1,1)(1,1).

Consider y=x32y=32x12y = x^{\frac{3}{2}} \Rightarrow y' = \frac{3}{2}x^{\frac{1}{2}}.
141+(32x12)2dx=141+94xdx=141+94x49d(1+94)=4923(1+(94x)32)14=127(80101313)\int^4_1 \sqrt{1+(\frac{3}{2}x^{\frac{1}{2}})^2} dx = \int^4_1 \sqrt{1+\frac{9}{4}x}dx = \int^4_1 \sqrt{1+\frac{9}{4}x} \frac{4}{9} d(1+\frac{9}{4}) = \frac{4}{9} \cdot \frac{2}{3}(1+(\frac{9}{4}x)^{\frac{3}{2}})|^4_1 = \frac{1}{27}(80 - \sqrt{10} - 13\sqrt{13})

dxdy=2y011+4y2=20114+y2dy\frac{dx}{dy} = 2y \Rightarrow \int^1_0 \sqrt{1+4y^2} = 2 \int^1_0 \sqrt{\frac{1}{4}+y^2} dy
Let y=12tanθdy=12sec2θdθy= \frac{1}{2}tan\theta \Rightarrow dy = \frac{1}{2}sec^2\theta d\theta.
20112secθ12sec2θdθ=12sec3θdθ=14(secθtanθ+lnsecθ+tanθ)0tan12=52+14ln(5+2)\Rightarrow 2\int^1_0 \frac{1}{2} sec \theta \cdot \frac{1}{2} sec^2\theta d\theta = \frac{1}{2}sec^3\theta d\theta \\= \frac{1}{4}(sec\theta tan\theta +ln| sec \theta +tan \theta |) |^{tan^{-1}2}_0 = \frac{\sqrt{5}}{2} + \frac{1}{4}ln(\sqrt{5}+2)

# Surface Area

Assume that ff' is continuous on [a,b][a,b] and gg' is continuous on [c,d][c,d].

  1. If the curve is y=f(x)y = f(x),axba \le x \le b and rotates about x-axis.
    S=2πf(x)ab1+(dydx)2dx=2πf(x)ab1+(f(x))2dxS = 2\pi f(x) \int^b_a \sqrt{1+(\frac{dy}{dx})^2} dx = 2\pi f(x) \int^b_a \sqrt{1+(f'(x))^2} dx
  2. If the curve is x=g(y)x = g(y),cxdc \le x \le d and rotates about y-axis.
    S=2πg(y)cd1+(dxdy)2dy=2πg(y)cd1+(g(y))2dyS = 2\pi g(y) \int^d_c \sqrt{1+(\frac{dx}{dy})^2} dy = 2\pi g(y) \int^d_c \sqrt{1+(g'(y))^2} dy

Find the surface area

  1. y=4x2,1x1y = \sqrt{4-x^2}, -1 \le x \le 1 rotate about x-axis.
  2. y=x2y = x^2, from (1,1)(1,1) to (2,4)(2,4) rotate about y-axis.

S=112π4x21+(2x24x2)2dx=8πS = \int^1_{-1}2 \pi \sqrt{4-x^2} \cdot \sqrt{1+(\frac{-2x}{2\sqrt{4-x^2}})^2} dx = 8\pi.
微積分的基本運算,我就略過囉!!

Consider x=yx = \sqrt{y} and dxdy=12y\frac{dx}{dy} = \frac{1}{2\sqrt{y}}.
S=142πy1+(12y)2dy=π6(171755)S = \int^4_1 2 \pi \sqrt{y} \sqrt{1 + (\frac{1}{2\sqrt{y}})^2}dy = \frac{\pi}{6}(17\sqrt{17} - 5\sqrt{5})
微積分的基本運算,我就略過囉!!


# Reference

  • 蘇承芳老師 - 微積分甲(一)109 學年度 - Calculus (I) Academic Year 109
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