LECTURE 3-1 AREA OF A REGION BOUNDED BY CURVES

Transcription

1 7 CALCULUS II DR. YOU 98 LECTURE 3- AREA OF A REGION BOUNDED BY CURVES If y = f(x) and y = g(x) are continuous on an interval [a, b] and f(x) g(x) for all x in [a, b], then the area of the region between the graphs of f (Top) and g (Bottom) on [a, b] is y y = f(x) Area = b f(x) dx a area under f(x) b = [f(x) g(x)] dx a b g(x) dx a area under g(x) a y = g(x) b x EXAMPLE Set up the definite integral that gives the area of the shaded region. (A) y = x (B) y = x x 6 (C) 6 (D) y = x 3 x y = x + 5 y = x 5 (E) y = x (F) y = x 5 5 x = y y = x

2 7 CALCULUS II DR. YOU 99 EXAMPLE EXAMPLE 3 Find the area of the region bounded by Find the area of the region bounded by (A) y = x, x =, x =, x-axis (A) y = x, x =, x =, x-axis (B) y = x 5, x =, x =, x-axis (B) y = x, x =, x =, x-axis

3 7 CALCULUS II DR. YOU EXAMPLE 3 EXAMPLE Find the area of the region bounded by Find the area of the region bounded by y = x + 6 and y = x y = x x and y = x

4 7 CALCULUS II DR. YOU EXAMPLE 5 EXAMPLE 6 Find the area of the region bounded by Find the area of the region bounded by y = x 5x + 6, y = x + 3x y = x 3 x + 6, y = x 3 + x

5 7 CALCULUS II DR. YOU EXAMPLE 7 EXAMPLE 8 Find the area of the region bounded by Find the area of the region bounded by y = x, y =, and y-axis. y = x, y =, and y-axis.

6 7 CALCULUS II DR. YOU 3 EXAMPLE 9 EXAMPLE Find the area of the region bounded by Find the area of the region bounded by y = x 3 3x + and y = x + y = x 3 x and y = x 3x

7 7 CALCULUS II DR. YOU EXAMPLE EXAMPLE Find the area of the region bounded by y = sin x on [, π] and x-axis Given the graphs of y = f(x) and y = f (x), find the area of the shaded region if f(x) dx = in the figure. y = f(x) y = f - (x)

8 7 CALCULUS II DR. YOU 5 If x = f(y) and x = g(y) are continuous on an interval [c, d] and f(y) g(y) for all x in [c, d], then the area of the region between the graphs of f (right) and g (left) on [c, d] is d Area = [f(y) g(y)] dy c x = g(y) y c d x = f(y) x EXAMPLE 3 EXAMPLE Find the area of the region bounded by Find the area of the region bounded by y = x, y = x, and x-axis. y = x and y = x + 6.

9 7 CALCULUS II DR. YOU 6 LECTURE 3- VOLUME: DISK METHOD DISK METHOD ABOUT x-axis If y = f(x) is continuous on an interval [a, b], the volume of the solid obtained by rotating the region under the curve f(x) from a to b about x-axis is The cylinder volume = [f(x i )] π x Then the volume of the solid: lim n n i= [f(x i )] π x b = [f(x)] π a dx DISK METHOD ABOUT y-axis Similarly, if x = g(y) is continuous on an interval [c, d], the volume of the solid obtained by rotating the region under the curve g(y) from c to d about y-axis is The cylinder volume = [f(y i )] π y The volume of the solid is lim n n i= [g(y)] π y d = [g(y)] π c dy

10 7 CALCULUS II DR. YOU 7 EXAMPLE Find the volume of the solid obtained by rotating the region bounded by y = x, x =, and x-axis about x-axis y = x 5 EXAMPLE EXAMPLE 3 Find the volume of the solid obtained by rotating the Find the volume of the solid obtained by rotating the region bounded by y = x, y =, x = in the region bounded by y = x, y =, x = in the quadrant I about x-axis. quadrant I about y-axis..

11 7 CALCULUS II DR. YOU 8 EXAMPLE EXAMPLE 5 Find the volume of the solid obtained by rotating the Find the volume of the solid obtained by rotating the region bounded by y = x, y = x, and x = about x-axis region bounded by y = x, y = x, and y = about y-axis

12 7 CALCULUS II DR. YOU 9 EXAMPLE 6 EXAMPLE 7 Find the volume of the solid obtained by rotating the Find the volume of the solid obtained by rotating the region bounded by y = x 3, y = 8, and y-axis about x-axis region bounded by y = x 3, y = 8, and y-axis about y-axis EXAMPLE 8 EXAMPLE 9 Find the volume of the solid obtained by rotating the Find the volume of the solid obtained by rotating the region bounded by y = x 3, y =, and x = about x-axis region bounded by y = x 3, y =, and x = about x-axis

13 7 CALCULUS II DR. YOU EXAMPLE EXAMPLE Find the volume of the solid obtained by rotating the Find the volume of the solid obtained by rotating the region bounded by y = x and y = x about x-axis. region bounded by y = x and y = x about y-axis. EXAMPLE EXAMPLE 3 Find the volume of the solid obtained by rotating the Find the volume of the solid obtained by rotating the region bounded by y = x (x ), y = x + and x = region bounded by y = x (x ), y = x + and x = about x-axis. about y-axis.

14 7 CALCULUS II DR. YOU EXAMPLE EXAMPLE 5 Find the volume of the solid obtained by rotating the Find the volume of the solid obtained by rotating the region bounded by y = e x, y = e and x = about x-axis. region bounded by y = e x, y = e and x = about y-axis. EXAMPLE 6 EXAMPLE 7 Find the volume of the solid obtained by rotating the Find the volume of the solid obtained by rotating the region bounded by y = x, x = and x = about x-axis. region bounded by y = x, x = and x = about y-axis.

15 7 CALCULUS II DR. YOU EXAMPLE 8 EXAMPLE 9 Find the volume of the solid obtained by rotating the Find the volume of the solid obtained by rotating the region bounded by y = x +, x =, x = 3 and y-axis region bounded by y = x + 3, y = 7, and y-axis about about y =. y =

16 7 CALCULUS II DR. YOU 3 EXAMPLE EXAMPLE Find the volume of the solid obtained by rotating the Find the volume of the solid obtained by rotating the region bounded by y = x 3, x =, and x-axis about x = region bounded by y = x, y-axis, and x-axis in. quadrant I about x =.

17 7 CALCULUS II DR. YOU EXAMPLE EXAMPLE 3 Find the volume of the solid obtained by rotating the Find the volume of the solid obtained by rotating the region bounded by y = x +, y =, and y-axis in region bounded by y = x, x =, and x-axis about y = quadrant I about y =. 8.

18 7 CALCULUS II DR. YOU 5 RIGHT CYLINDER VOLUME Let A(x i ) be the area of the cross section at x = x i b = A(x) dx a V = lim n n i= A(x i ) x i EXAMPLE EXAMPLE 5 The solid lies between planes perpendicular to the x-axis The solid lies between planes perpendicular to the y-axis at x. The cross-sections perpendicular to the xaxis are square whose sides run from y = x + to y = axis are square whose side run from x = y to x = 3 at y. The cross-sections perpendicular to the y- x. Find the volume of this solid. y. Find the volume of this solid.

19 7 CALCULUS II DR. YOU 6 EXAMPLE 6 EXAMPLE 7 The solid lies between planes perpendicular to the x-axis The solid lies between planes perpendicular to the x-axis at x. The cross-sections perpendicular to the xaxis are semi-circle whose diameters run from y = x to axis are semi-circle whose diameters run from y = x to at x. The cross-sections perpendicular to the x- y = x. Find the volume of this solid. y = x. Find the volume of this solid.

20 7 CALCULUS II DR. YOU 7 LECTURE 3-3 ARC LENGTHS AND SURFACE AREA OF REVOLUTION THE ARC LENGTH of a smooth curve y = f(x) on the interval [a, b]: We divide the interval [a, b] into n subintervals of length x = b a, where x n k is the right endpoint of the kth subinterval and p k is the right point on the graph of y = f(x), for k =,,, n. ) Each distance from p i to p i is = ( x) + ( y) (by Pythagorean theorem; x = x i x i, y = f(x i ) f(x i )) = + ( y x ) x = + (f (c i )) x (by Mean Value theorem, y = f(c i ) x for some c i in [x i, x i ] ) ) Assume that L n = The sum of all distances from p i to p i The arc length s = lim n L n n = lim { + (f (c i )) x} n i= b = + [f (x)] dx a THE ARC LENGTH of a smooth curve x = g(y) on the interval [c, d]: d Similarly, the arc length s = + [g (y)] c dy EXAMPLE Find the length of curve which is graph of the function y = 3 x3 on x.

21 7 CALCULUS II DR. YOU 8 EXAMPLE EXAMPLE 3 Find the arc length of the graph of y = x3 interval [, ]. + x on the Find the arc length of the graph of y = x3 + on the 6 x interval [, ].

22 7 CALCULUS II DR. YOU 9 EXAMPLE EXAMPLE 5 Find the arc length of the graph of y = (x + ) 3/ + 3 on the interval [, 3]. Find the arc length of the graph of y = x3 + on the 3 x interval [, 3].

23 7 CALCULUS II DR. YOU EXAMPLE 6 EXAMPLE 7 Find the arc length of the graph of (y ) 3 = x on the Find the arc length of the graph of y = ln(cos x) on the interval [, 8]. interval [, π ].

24 7 CALCULUS II DR. YOU THE SURFACE AREA by rotating the smooth curve y = f(x) on the interval [a, b]: We divide the interval [a, b] into n subintervals of length x = b a, where x n k is the right endpoint of the kth subinterval and p k is the right point on the graph of y = f(x), for k =,,, n. ) Each surface area by rotating the curve y = f(x) from p i to p i is πf(x i ) ( x) + ( y) ( x = x i x i, y = f(x i ) f(x i )) = πf(x i ) + ( y x ) x = πf(x i ) + (f (c i )) x (by Mean Value theorem, y = f(c i ) x for some c i in [x i, x i ] ) ) Assume that S n = The sum of all surface areas by rotating the curve y = f(x) from p i to p i 3) The surface S which is obtained by rotating the curve; y = f(x) on [a, b] about x-axis is = lim n S n n = lim {πf(x i ) + (f (c i )) x} n i= b = πf(x) + (f (x)) dx a THE SURFACE AREA of rotating the smooth curve x = g(y) on the interval [c, d]: Similarly, the surface S which is obtained by rotating the curve x = g(y) on [c, d] about y-axis is d πg(y) + [g (y)] dy c

25 7 CALCULUS II DR. YOU EXAMPLE 8 EXAMPLE 9 Find the area of the surface obtained by rotating the region Find the area of the surface obtained by rotating the region bounded by f(x) = x on [,] about x-axis. bounded by f(x) = x on [,] about x-axis. EXAMPLE EXAMPLE Find the area of the surface obtained by rotating the region bounded by f(x) = tan x on [, π ] about x-axis. Find the area of the surface obtained by rotating the region bounded by f(x) = x on [,] about x-axis.

26 7 CALCULUS II DR. YOU 3 LECTURE 3- WORK DEFINITION: ) If an object is moved a distance D in the direction of an applied constant force F, then the work W done by the force is defined as W = FD. ) In the SI metric, the mass is required in kilograms, the displacement is meter, the time in seconds, and the force in newtons (N = kg m/s )) 3) Gravity Force is F = mg m is mass and g = 9.8 m/s THEOREM: ) If an object is moved along a straight line by a continuously varying force F(x), then the work W done by the force as object is moved from x = a and x = b is given by b W = F(x) dx a ) Newton s second law: If an object is moved along a straight line with position function s(t), then the force F on the object is defined by F = m d s dt where m is its mass (mass = density volume) 3) Hook s Law: The force F required to compress or stretch a spring is proportional to the distance d that the spring is compressed or stretched from its original length. That is F = kd where k is called a force constant EXAMPLE EXAMPLE The magnitude of the required force F is the weight of the A force of 75 pounds compresses a spring 3 inches from object. Find the work done in lifting the object feet. its natural length of 5 inches. Find the work done in compressing the spring an additional 3 inches

27 7 CALCULUS II DR. YOU EXAMPLE 3 EXAMPLE Find the work required to compress a spring from its A spring has a natural length of meter. A force of N natural length of ft. to a length.75 ft. if the force holds the spring stretched to a total length of.8 meter. constant is k = 6 lb/ft. (A) Find the force constant k. (B) How much work will it take to stretch the spring meters beyond its natural length?

28 7 CALCULUS II DR. YOU 5 EXAMPLE 5 EXAMPLE 6 A force of N is required to hold a spring that has been A tank has the shape of an inverted circular cone with stretched from its natural length of cm to a length of 5 height m and base radius m. It is filled with water to cm. How much work is done in stretching the spring from a height of 8 m. Find the work required to empty the tank 5 cm to 8 cm? by pumping all the water to the top of the tank (the density of water is kg/m 3.

29 7 CALCULUS II DR. YOU 6 PRACTICE PROBLEMS FOR UNIT 3. Set up integral which represents the area of the shaded region. (A) (B) 3 y = x + y = 3 x y = x y = x 3 (C) (D) y = x y = 8 x y = x 3 y = x (E) (F) x = y y y = x x = y x = y y. Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x and y. Find the area of the region. (A) y = x x and y = x. (B) y = x 3 and y = 9x. (C) x + y = and x = y (D) x = y(y ) and y = x 3 (E) y = cos x, y = cos x and x π (F) y = x sin x ( x π) and x-axis.

30 7 CALCULUS II DR. YOU 7 3. Let R be the region bounded by the following function. Find the volume of the solid by revolving the region R about the given line. (A) y = x, y =, x = ; about x-axis (B) y = cos x ( x π ) x = ; about x-axis (C) y = x, y = x; about x-axis (E) y = x, y = x, y = 6; about y-axis (G) y = x +, x =, y = ; about x = (I) y = x (x ), y =, y-axix; about y = 6 (D) y = x, y = x x 3 ; about x-axis (F) y = x, y =, y =, x = ; about y-axis (H) y = x, y =, x = ; about x = (J) y = x, x = y ; about y =. Use revolution of a region to find the volume of a sphere of radius r. 5. The solid whose base is the region bounded by y = x and y = and whose cross sections perpendicular to xaxis are square. 6. Find the arc length of the following curves and (A) y = 3 (x )3/, x (B) y = x 3/, x (C) y = x + 8x, x x3 (D) y = +, x 3 x 7. find the surface area obtained by revolving the region bounded by the curves (A) y = x +, x =, x =, x-axis (B) y = x3 +, x =, x =, x-axis 3 x 8. A force of 3 N is required to maintain a spring stretched from its natural length of cm to length of 5 cm. How much work is done in stretching the spring from cm to cm?

31 7 CALCULUS II DR. YOU 8 SOLUTIONS. (A) x 3 x (B) 3 x (x + ) (C) x (D) y + y 3/ dx + x x3 dx dx + 8 x dx dx = x x + dx dy or 8 x/3 3 (E) y y (y y ) (F) y (y 3) 3. (A) (x x ) x (B) (C) 3 6 x 3 9x (3 y ) y (D) 3y y(y ) π (E) x sin x dx 8 x dx dy = 3 y 6y dy dy = 3 y y dy 3 dx = x x dx = dx + 9x x 3 dx = 8 dx dy = 6 3 dy = y y dy = 3 3 π π 3. (A) π ( x ) dx = 5 π 5 (B) π π/ (C) xπ (D) 6x π 6 (E) y π cos x dx = π x sin x dx = π + 3π = π π/ +cos(x) dx x πdx = 3π 3 dx = π dx (x x3 ) π dx = 56π 35 dy 6 ( y ) π dy = 5π (F) (y + ) π dy = 6π 5 (G) Shift left unit: y = x +, x = x = y, x = (y ) πdy ( ) π dy = 8π 5 (H) Shift right unit: y = x, x = x = y +, x = π dy (y + ) dy = 3π 5 (I) Shift down 6 units: y = x, y = y = x 6, y = (x 6) π dx ( ) π dx = 38.π (J) Shift up unit: y = x, y = x y = x +, y = x + ( x + ) π dx (x + ) π dx = 9π 3

32 7 CALCULUS II DR. YOU 9. The sphere is same the solid by revolving the semicircle y = r x about x-axis on r x r r π (r x ) r dx = π (r x r 3 x3 ) = r 3 r3 π 5. The length of a side of each rectangle is x on x. 6. (A) y = x ; + (y ) = x The are length is x (B) y = 3 x; + (y ) = + 9 x The are length is + 9 x (C) y = x 3 x 3 ; ( x ) dx = 6 5 dx = 9 u du = 5 3 dx = 9 u du = 8 7 ( ) u = + 9 x + (y ) = + (x 3 x 3) = + x 6 + 6x 6 = x x 6 = (x 3 + x 3) The are length is x 3 + x 3 dx = 3 3 (D) y = x x ; + (y ) = + (x x ) = x + x The are length is x + x dx = (A) π(x + ) + dx = 8 5π (B) By 6 (D), π ( x3 3 + x ) (x + x ) dx = π 8. By Hook s Law, F = kx 3 = k(.3) k = N/m F =.8 = 8 N