Overview Basic Guidelines Slicing –Circles –Squares –Diamonds –Triangles Revolving Cylindrical Shells

Basic Guidelines Sketch the solid and a typical cross-section Find a formula for A(x) Find the limits of the integration Integrate A(x) to find the volume

Slicing (Circles) The object lies between x=0, x=4, so the limits of integration will be from zero to four. The area of on circular cross section is Πr 2 The radius r = f(x) = √x. A(x)= Π (√x) 2 = Πx The volume is the integration of the area giving us 0 ∫ 4 A(x) dx = Π 0 ∫ 4 x dx Π [ x 2 /2] 0 4 = Π ( 16/2 – 0) = 8Π

Cross Section

Slicing (Square) The limits of integration is x= -1, x=1 The area of a square base cross section is s 2 The length of one side is 2f(x), giving us (2f(x)) 2, f(x)= √1-x 2 4(1-x 2 ) = 4-4x 2 Volume= -1 ∫ 1 A(x)dx = -1 ∫ 1 4-4x 2 dx = 2 0 ∫ x 2 = 8 0 ∫ 1 1-x 2 = 8[x-(x3/3)] 0 1 = 16/3

Cross Section

Slicing (Diamond) The cross sections of this object are squares with diagonals in the xy plane The area of a square is s 2, f(x)=√x The side of each square is (√2 f(x)) (√2 f(x)) 2 = 2x The limits of integration are x=0, x=4, thus 0 ∫ 4 2x dx = [x 2 ] 0 4 = 16-0= 16.

Cross Section

Slicing (Triangle) The cross sections of this object are equilateral triangles with bases in the xy plane. the limits of integration are x=0, x=4. The area of a triangle is 0.5bh The base of the triangle is 2 f(x), the height is √3 f(x). (2 √x) (√3 √x) (0.5)= √3x = A(x) Volume= 0∫4 √3x = √3 0∫4x = √3[ x2/2] 0 4 = 8 √3

Cross Section

Revolving Revolving this shape around the x-axis gives a heresy kiss shape with a circular cross section between x=0, x=2. Again the area is Π r 2 The radius, r= f(x)= x 3 A(x)=Π (x 3 ) 2 = x 6 The volume is the integration of the area giving Π 0 ∫ 2 x 6 dx= Π [x 7 /7] 0 2 = (128/7)Π

Washer Cross Section Revolving this area about the x-axis produces a washer cross section. Area of a washer= area of larger circle –area of smaller circle. Area if large = Π (2x) 2 Area of small = Π x 2 4x 2 Π – x 2 Π = Π (4x 2 -x 2 ) = 3x 2 Π The limit of integration is between x=0, x=1, thus 3Π 0 ∫ 1 x 2 = 3Π [ x3/3] 0 1 = (1/3)x 3 Π= Π

Cross Section

Rotations (Vase) The region between the graph of 2 + x cos (x) and the x-axis is rotated over the interval [-2, 2] about the x-axis The cross section is circular so the area is A(x) = Πr2, the radius being f(x). The volume = 0 ∫ 2 2A(x) dx Use NINT to integrate NINT(Π(2 + x cos x) 2, x,-2,2) = 52.43

Cylindrical Shells (Bunt Cake) Region enclosed by y=3x-x 2, revolved about the y-axis. Limits of integration from x=0, x=3. Lateral area of cylinder is 2Πrh The r=x, the h= f(x)= 3x-x 2 (2Πx)( 3x-x 2 )=lateral area; volume =integration 2Π 0 ∫ 3 3x 2 -x 3 = 2Π [x3-x4/4] 0 3 = 2Π (27- (81/4)) = 54/4 Π =13.5Π

Cross Section