Chapter 6     

Section 1

          ~The area A of the region bounded by the curves y=f(x), y=g(x), and the lines x=a, x=b, where f and g are continuous and f(x)  g(x) for all x in [a,b] is                 A=

          A= or A=

          Ex.: Find the area bounded by the curves:

Y= 20-x²          y = x²-12                                 

        Set equations equal to each and solve for x to find interval of the integral.

            20 - x² = x² – 12                                              A = [(32(4)-2(4³ ))-(32(-4)-2(-4³))]

            32 - x² = x²                                                                        A = [(128 - 128/3) - (-128 + 128/3)]

            32 = 2x²                                                                              A = [(128 – 128/3 + 128 – 128/3)]

            x² = 16                                                             Find a common denominator

            x = 4 (a = -4, b = 4)                                   A = 384/3 –128/3 + 384/3 – 128/3

          So if                                        A = -256/3 + 768/3 = 512/3

          Then                          Answer = A = 512/3

          Ex:           Find the area enclosed by the line y = x – 1 and parabola y² = 2x + 6

           Solve for x

            Y = x – 1         y² = 2x + 6

            X = y + 1         x = ½(y²) – 3

           Integrate between appropriate y-values

            Y= -2 and y = 4

            = [-1/2(y³/3)+(y²/2)+4y]

            = [(-1/6(4³) + 4²/2 + 4(4)) – (-1/6(2³ + 2²/2 + 4(2)))]

            = -1/6(64) + 8 + 16 – (4/3 + 2 – 8) = 18

Answer = A = 18 units²

          Trigonometric Functions:

          D/dx(sin x) = cos x

          D/dx(cos x) = -sin x

          D/dx(tan x) = sec²x

          Ex: Find the area enclosed by y = sin x and y = -cos x , x = 0  x =

            Solve for x

          Sin x = -cos x

          Sin x/cos x = -1

          Tan x = -1

          X = 3/4

          Integrate between appropriate x-values

          (x = 0 to x = 3/4),(x = 3/4 to x =)

Section 2

           Volume = V = AH

            a. Cylinder                                            b. circular cylinder                         c. Rectangular box

           V = AH                                                  V =  r²h                                              V = LWH

          Cross Sections:

            For a solid S that isn’t a cylinder we first “cut” s into pieces and approximate each piece by a cylinder. We estimate the volume of S by adding the                  volumes of the cylinders. We then arrive at the exact volume of S though a limiting process in which the number of pieces becomes large.

            We start by intersection S with a plane and obtaining a plane region that is called a cross-section. Let A(x) be the area of the cross-section of S in a                plane Px perpendicular to the x-axis and passing through the point x, where . The cross-sectional area A(x) will vary as x increases from a to

          *Definition of Volume

            If the cross-sectional area of S in the plane Px, through x and perpendicular to the x-axis is A(x) where A is a continuous function, then the volume of s            is , where A(x) is area.

         Ex:

          Cross sections perpendicular to the x-axis are squares

                     solve for

à Solution

Disc

           Ex: Find the volume of the solid obtained by rotating about the x-axis the region under the curve  from 0 to 1.

          Find A(x) = () from 0 to 1

dx

                     xdx

                    (1/2)- (0/2)

           Solution: V(x) = /2

Washer

           Ex: Find the volume of the resulting solid from the region enclosed by the curves y = x and y = x is rotated about x-axis, 0 to 1

ccc
cccc

= ()

     V =