Area of a Region Between Two Curves Volume Arc Length # Area of a Region Between Two CurvesThe area A of the region bounded by the curves y = f ( x ) , y = g ( x ) y =f(x), y= g(x) y = f ( x ) , y = g ( x ) and the lines x = a , x = b x=a, x=b x = a , x = b , where f f f and g g g are continuous ,is A = ∫ a b ∣ f ( x ) − g ( x ) ∣ d x A = \int^b_a |f(x) - g(x)|dx A = ∫ a b ∣ f ( x ) − g ( x ) ∣ d x
# Volume# The Disk MethodV = ∫ a b A ( x ) d x V = \int^b_a A(x)dx V = ∫ a b A ( x ) d x (about x-axis)V = ∫ c d A ( y ) d y V = \int^d_c A(y)dy V = ∫ c d A ( y ) d y (about y-axis)# The Shell MethodV = ∫ a b 2 π x f ( x ) d x V = \int^b_a 2 \pi xf(x) dx V = ∫ a b 2 π x f ( x ) d x (about y-axis)V = ∫ c d 2 π y f ( y ) d y V = \int^d_c 2 \pi yf(y) dy V = ∫ c d 2 π y f ( y ) d y (about x-axis)Find the volume
by rotating about the y-axis the region bounded by y = 2 x 2 − x 3 y = 2x^2 - x^3 y = 2 x 2 − x 3 and y = 0 y = 0 y = 0 .
V = ∫ 0 2 ( 2 π x ) ( 2 x 2 − x 3 ) d x = 2 π ∫ 0 2 2 x 3 − x 4 d x = 2 π 1 2 x 4 − 1 5 x 5 ∣ 0 2 = 2 π ( 8 − 32 5 ) = 16 5 π V = \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 V = ∫ 0 2 ( 2 π x ) ( 2 x 2 − x 3 ) d x = 2 π ∫ 0 2 2 x 3 − x 4 d x = 2 π 2 1 x 4 − 5 1 x 5 ∣ 0 2 = 2 π ( 8 − 5 3 2 ) = 5 1 6 π
# Arc LengthP i P i + 1 ‾ = ( x i + 1 − x i ) 2 + ( f ( x i + 1 ) − f ( x i ) ) 2 = ( Δ x ) 2 + ( Δ y ) 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} P i P i + 1 = ( x i + 1 − x i ) 2 + ( f ( x i + 1 ) − f ( x i ) ) 2 = ( Δ x ) 2 + ( Δ y ) 2 ⇒ ∑ i = 1 n P i P i + 1 ‾ = ∑ i = 1 n ( Δ x ) 2 + ( Δ y ) 2 = ∑ i = 1 n ( ( Δ x ) 2 ( Δ x ) 2 ) + ( Δ y ) 2 ( Δ x ) 2 Δ x = ∫ x i x n 1 + ( f ′ ( x ) ) 2 d x \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 ⇒ ∑ i = 1 n P i P i + 1 = ∑ i = 1 n ( Δ x ) 2 + ( Δ y ) 2 = ∑ i = 1 n ( ( Δ x ) 2 ( Δ x ) 2 ) + ( Δ x ) 2 ( Δ y ) 2 Δ x = ∫ x i x n 1 + ( f ′ ( x ) ) 2 d x
Assume that f ′ f' f ′ is continuous on [ a , b ] [a,b] [ a , b ] and g ′ g' g ′ is continuous on [ c , d ] [c,d] [ c , d ] .
If the arc is described by y = f ( x ) y=f(x) y = f ( x ) ,a ≤ x ≤ b a \le x \le b a ≤ x ≤ b then L = ∫ a b 1 + ( d y d x ) 2 d x = ∫ a b 1 + ( f ′ ( x ) ) 2 d x L = \int^b_a \sqrt{1+(\frac{dy}{dx})^2}dx = \int^b_a \sqrt{1+(f'(x))^2}dx L = ∫ a b 1 + ( d x d y ) 2 d x = ∫ a b 1 + ( f ′ ( x ) ) 2 d x If the arc is described by x = g ( y ) x=g(y) x = g ( y ) ,c ≤ y ≤ d c \le y \le d c ≤ y ≤ d then L = ∫ c d 1 + ( d y d x ) 2 d y = ∫ c d 1 + ( g ′ ( y ) ) 2 d y L = \int^d_c \sqrt{1+(\frac{dy}{dx})^2}dy = \int^d_c \sqrt{1+(g'(y))^2}dy L = ∫ c d 1 + ( d x d y ) 2 d y = ∫ c d 1 + ( g ′ ( y ) ) 2 d y y 2 = x 3 y^2=x^3 y 2 = x 3 ,Find the length of arc
from ( 1 , 1 ) (1,1) ( 1 , 1 ) to ( 4 , 8 ) (4,8) ( 4 , 8 ) .y 2 = x y^2 = x y 2 = x ,Find the length of arc
from ( 0 , 0 ) (0,0) ( 0 , 0 ) to ( 1 , 1 ) (1,1) ( 1 , 1 ) .Consider y = x 3 2 ⇒ y ′ = 3 2 x 1 2 y = x^{\frac{3}{2}} \Rightarrow y' = \frac{3}{2}x^{\frac{1}{2}} y = x 2 3 ⇒ y ′ = 2 3 x 2 1 .∫ 1 4 1 + ( 3 2 x 1 2 ) 2 d x = ∫ 1 4 1 + 9 4 x d x = ∫ 1 4 1 + 9 4 x 4 9 d ( 1 + 9 4 ) = 4 9 ⋅ 2 3 ( 1 + ( 9 4 x ) 3 2 ) ∣ 1 4 = 1 27 ( 80 − 10 − 13 13 ) \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}) ∫ 1 4 1 + ( 2 3 x 2 1 ) 2 d x = ∫ 1 4 1 + 4 9 x d x = ∫ 1 4 1 + 4 9 x 9 4 d ( 1 + 4 9 ) = 9 4 ⋅ 3 2 ( 1 + ( 4 9 x ) 2 3 ) ∣ 1 4 = 2 7 1 ( 8 0 − 1 0 − 1 3 1 3 )
d x d y = 2 y ⇒ ∫ 0 1 1 + 4 y 2 = 2 ∫ 0 1 1 4 + y 2 d y \frac{dx}{dy} = 2y \Rightarrow \int^1_0 \sqrt{1+4y^2} = 2 \int^1_0 \sqrt{\frac{1}{4}+y^2} dy d y d x = 2 y ⇒ ∫ 0 1 1 + 4 y 2 = 2 ∫ 0 1 4 1 + y 2 d y Let y = 1 2 t a n θ ⇒ d y = 1 2 s e c 2 θ d θ y= \frac{1}{2}tan\theta \Rightarrow dy = \frac{1}{2}sec^2\theta d\theta y = 2 1 t a n θ ⇒ d y = 2 1 s e c 2 θ d θ .⇒ 2 ∫ 0 1 1 2 s e c θ ⋅ 1 2 s e c 2 θ d θ = 1 2 s e c 3 θ d θ = 1 4 ( s e c θ t a n θ + l n ∣ s e c θ + t a n θ ∣ ) ∣ 0 t a n − 1 2 = 5 2 + 1 4 l n ( 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) ⇒ 2 ∫ 0 1 2 1 s e c θ ⋅ 2 1 s e c 2 θ d θ = 2 1 s e c 3 θ d θ = 4 1 ( s e c θ t a n θ + l n ∣ s e c θ + t a n θ ∣ ) ∣ 0 t a n − 1 2 = 2 5 + 4 1 l n ( 5 + 2 )
# Surface AreaAssume that f ′ f' f ′ is continuous on [ a , b ] [a,b] [ a , b ] and g ′ g' g ′ is continuous on [ c , d ] [c,d] [ c , d ] .
If the curve is y = f ( x ) y = f(x) y = f ( x ) ,a ≤ x ≤ b a \le x \le b a ≤ x ≤ b and rotates about x-axis.S = 2 π f ( x ) ∫ a b 1 + ( d y d x ) 2 d x = 2 π f ( x ) ∫ a b 1 + ( f ′ ( x ) ) 2 d x S = 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 S = 2 π f ( x ) ∫ a b 1 + ( d x d y ) 2 d x = 2 π f ( x ) ∫ a b 1 + ( f ′ ( x ) ) 2 d x If the curve is x = g ( y ) x = g(y) x = g ( y ) ,c ≤ x ≤ d c \le x \le d c ≤ x ≤ d and rotates about y-axis.S = 2 π g ( y ) ∫ c d 1 + ( d x d y ) 2 d y = 2 π g ( y ) ∫ c d 1 + ( g ′ ( y ) ) 2 d y S = 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 S = 2 π g ( y ) ∫ c d 1 + ( d y d x ) 2 d y = 2 π g ( y ) ∫ c d 1 + ( g ′ ( y ) ) 2 d y Find the surface area
y = 4 − x 2 , − 1 ≤ x ≤ 1 y = \sqrt{4-x^2}, -1 \le x \le 1 y = 4 − x 2 , − 1 ≤ x ≤ 1 rotate about x-axis.y = x 2 y = x^2 y = x 2 , from ( 1 , 1 ) (1,1) ( 1 , 1 ) to ( 2 , 4 ) (2,4) ( 2 , 4 ) rotate about y-axis.S = ∫ − 1 1 2 π 4 − x 2 ⋅ 1 + ( − 2 x 2 4 − x 2 ) 2 d x = 8 π S = \int^1_{-1}2 \pi \sqrt{4-x^2} \cdot \sqrt{1+(\frac{-2x}{2\sqrt{4-x^2}})^2} dx = 8\pi S = ∫ − 1 1 2 π 4 − x 2 ⋅ 1 + ( 2 4 − x 2 − 2 x ) 2 d x = 8 π . 微積分的基本運算,我就略過囉!!
Consider x = y x = \sqrt{y} x = y and d x d y = 1 2 y \frac{dx}{dy} = \frac{1}{2\sqrt{y}} d y d x = 2 y 1 .S = ∫ 1 4 2 π y 1 + ( 1 2 y ) 2 d y = π 6 ( 17 17 − 5 5 ) 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}) S = ∫ 1 4 2 π y 1 + ( 2 y 1 ) 2 d y = 6 π ( 1 7 1 7 − 5 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