Chapter 7: Applications of Integration

Transcription

1 Chapter 7: Applications of Integration Fall 214 Department of Mathematics Hong Kong Baptist University 1 / 21

2 7.1 Volumes by Slicing Solids of Revolution In this section, we show how volumes of certain three-dimensional regions (or solids) can be expressed as definite integrals and thereby determined. Key idea: Volumes by Slicing. By dividing solids into thin slices by parallel planes, we determine the volumes for each slide and then sum these volumes to find the volume of the solid. 2 / 21

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4 Suppose that the solid S lies between planes perpendicular to the x-axis at positions x = a and x = b, and that the cross-sectional area of S in the plane perpendicular to the x-axis at x is a known function A(x), for a x b. Assume that A(x) is continuous on [a, b]. Let a = x < x 1 < < x n 1 < x n = b. This results in n slices of which the ith has thickness x i = x i x i 1. By the Intermediate-Value theorem, the ith slice has volume V i = A(c i ) x i for some c i in [x i 1, x i ]. The volume of the solid is then V = n V i = i=1 n A(c i ) x i. i=1 4 / 21

5 Letting n approach infinity in such a way that max x i approaches, we obtain the definite integral of A(x) over [a, b] as the limit of the above Riemann sum. Theorem The volume V of a solid between x = a and x = b having cross-sectional area A(x) at position x is V = b a A(x)dx. 5 / 21

6 Solids of Revolution Many common solids have circular cross-sections in planes perpendicular to some axis. Such solids are called solids of revolution because they can be generated by rotating a plane region about an axis in that plane so that it sweeps out the solid. If the region R bounded by y = f (x), y =, x = a and x = b is rotated about the x-axis, then the cross-sections of the solid generated are circular disks of radius f (x). This leads to A(x) = π(f (x)) 2 and the volume of the solid of revolution as V = π b a (f (x)) 2 dx. 6 / 21

7 Example 1: Let R be the region bounded by y = 4 x 2 and y =. Find the volume of the solid obtained by revolving R about the y-axis. 4 y x = 4 y y x 2 x 7 / 21

8 Solution: The area of ring revolved by the line between 4 y and 4 y about the y-axis is A(y) = πx 2 x= 4 y = π(4 y) The volume of the revolving solid is V = = 4 4 A(y)dy π(4 y)dy = (4πy π 2 y 2 ) = 8π. 4 8 / 21

9 Example 2: Let R be the region bounded by y = 4 x 2 and y =. Find the volume of the solid obtained by revolving R about the line x = 3. 4 y x = 4 y y x 2 3 x 9 / 21

10 Solution: The area of ring revolved by the line between 4 y and 4 y about x = 3 is A(y) = π(3 + 4 y ) 2 π(3 4 y ) 2 = 12π 4 y. The volume of the revolving solid is V = = 4 4 A(y)dy 12π 4 ydy ( ) 3 4 = 8π 4 y = 64π. 1 / 21

11 Example 3: Find the volume of a solid ball having radius a. Solution: Note that the ball can be generated by rotating the half disk, y a 2 x 2, a x a, about the x-axis. Therefore, the volume is V = a = 2π a a π( a 2 x 2 ) 2 dx (a 2 x 2 )dx = 2π(a 2 x x 3 3 ) a = 4 3 πa3. 11 / 21

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13 Example 4: Find the volume of the infinitely long horn that is generated by rotating the region bounded by y = 1/x and y = and lying to the right of x = 1 about the x-axis. Solution: The volume of the horn is V = 1 = π lim R π( 1 x )2 dx R = π lim R 1 x 2 dx 1 R x 1 = π lim R ( 1 R 1) = π / 21

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15 7.3 Arc Length and Surface Area (optional) Let f be a function defined on [a, b] and having a continuous derivative f there. We divide the internal [a, b] into n subintervals with the partition, {a = x < x 1 < < x n = b}. The arc length in [x k 1, x k ] may be approximated by ( x k ) 2 + ( y k ) 2. The arc length of the curve in [a, b] is then approximated by L n = n k=1 ( x k ) 2 + ( y k ) / 21

16 y y = f (x) x k y k x a x 1 x k 1 x k x n b x 16 / 21

17 Letting n in such a way that max( x i ), the arc length of the curve is given by s = lim L n = lim n n = lim = n ( xk ) 2 + ( y k ) 2 k=1 n n k=1 b a 1 + ( y k x k ) 2 x k 1 + ( dy dx )2 dx. In other words, L n is a Riemann sum for the arc length s of the curve y = f (x) from x = a to x = b, b s = 1 + ( dy b dx )2 dx = 1 + (f (x)) 2 dx. a a 17 / 21

18 Example 5: Find the length of the curve y = x 2/3 from x = 1 to x = 8. Solution: Note that dy/dx = (2/3)x 1/3 is continuous between x = 1 and x = 8. The arc length of the curve is 8 s = 1 + ( dy 8 dx )2 dx = x 2/3 dx = x 1/3 9x 2/3 + 4dx. Let u = 9x 2/ Then du = 6x 1/3 dx and s = u 1/2 du = u3/ = / 21

19 Example 6: Find the length of the curve y = x from x = 1 to x = 2. 32x 2 Solution: Note that 1 + ( dy dx )2 = 1 + (4x x 3 )2 = (4x x 3 )2. The arc length of the curve is 2 s = 1 + ( dy dx )2 dx = 1 2 = (x x 2 ) 2 = (4x x 3 )dx 19 / 21

20 Example 7: Find the arc length of a circle with a radius of a. Solution: Consider t < π/2. We have y = a 2 x 2. The arc length of the circle is given by a ( ) 2 dy a x s = dx = ( dx a2 x 2 )2 dx = 4 a a a2 x 2 dx. Method 1: Let z = x/a. We have 1 1 s = 4a dz = 4a arcsin(z) 1 = 2πa. 1 z 2 2 / 21

21 Method 2: Let x = a sin(t), then dx = a cos(t)dt. We have s = 4 = 4 = 4a a π/2 a a 2 x 2 dx π/2 = 2πa. a a 2 a 2 sin 2 (t) 1dt a cos(t)dt 21 / 21