VolumesCross SectionsSlide 3Slide 4Try It OutRevolving a FunctionDisksSlide 8Try It Out!WashersApplicationRevolving About y-AxisSlide 13Flat WasherAssignmentSlide 16Slide 17Shell MethodThe ShellSlide 20Hints for Shell MethodRotation About x-AxisRotation About Non-coordinate AxisSlide 24Slide 25Slide 26Slide 27VolumesLesson 6.2Cross Sections•Consider a square at x = c with side equal to side s = f(c)•Now let this be a thinslab with thickness Δx•What is the volume of the slab?•Now sum the volumes of all such slabscf(x)( )21( )niib af xn=-�baCross Sections•This suggests a limitand an integralcf(x)( )21( )niib af xn=-�ba( ) ( )2 21lim ( ) ( )bniniab af x f x dxn��=-=��Cross Sections•We could do similar summations (integrals) for other shapesTrianglesSemi-circlesTrapezoidscf(x)baTry It Out•Consider the region bounded above by y = cos xbelow by y = sin xon the left by the y-axis•Now let there be slices of equilateral triangles erected on each cross section perpendicular to the x-axis•Find the volumeRevolving a Function•Consider a function f(x) on the interval [a, b]•Now consider revolvingthat segment of curve about the x axis•What kind of functions generated these solids of revolution?f(x)abDisks•We seek ways of usingintegrals to determine thevolume of these solids•Consider a disk which is a slice of the solidWhat is the radiusWhat is the thicknessWhat then, is its volume?dxf(x)[ ]2Volume of slice = ( )f x dxpDisks•To find the volume of the whole solid we sum thevolumes of the disks•Shown as a definite integralf(x)ab[ ]2( )baV f x dxp=�Try It Out!•Try the function y = x3 on the interval 0 < x < 2 rotated about x-axisWashers•Consider the area between two functions rotated about the axis•Now we have a hollow solid•We will sum the volumes of washers•As an integralf(x)abg(x)[ ] [ ]{ }2 2( ) ( )baV f x g x dxp= -�Application•Given two functions y = x2, and y = x3Revolve region between about x-axisWhat will be the limits of integration?What will be the limits of integration?( ) ( )12 22 30V x x dxp� �= -� �� ��Revolving About y-Axis•Also possible to revolve a function about the y-axisMake a disk or a washer to be horizontal•Consider revolving a parabola about the y-axisHow to represent the radius?What is the thicknessof the disk?Revolving About y-Axis•Must consider curve asx = f(y)Radius = f(y)Slice is dy thick•Volume of the solid rotatedabout y-axis[ ]2( )baV f y dyp=�Flat Washer•Determine the volume of the solid generated by the region between y = x2 and y = 4x, revolved about the y-axisRadius of inner circle?•f(y) = y/4Radius of outer circle?• Limits?•0 < y < 16( )f y y=Assignment•Lesson 6.2A•Page 372•Exercises 1 – 29 odd 2410 169y x x  Find the volume generated when this shape is revolved about the y axis.We can’t solve for x, so we can’t use a horizontal slice directly. 2410 169y x x  If we take a vertical sliceand revolve it about the y-axiswe get a cylinder.Shell Method•Based on finding volume of cylindrical shellsAdd these volumes to get the total volume•Dimensions of the shellRadius of the shellThickness of the shellHeightThe Shell•Consider the shell as one of many of a solid of revolution•The volume of the solid made of the sum of the shellsf(x)g(x)xf(x) – g(x)dx[ ]2 ( ) ( )baV x f x g x dxp= -�Try It Out!•Consider the region bounded by x = 0, y = 0, and 28y x= -2 2202 8V x x dxp= � -�Hints for Shell Method•Sketch the graph over the limits of integration•Draw a typical shell parallel to the axis of revolution•Determine radius, height, thickness of shell•Volume of typical shell•Use integration formula 2 radius height thicknessp�� � �2baVolume radius height thicknessp= � � ��Rotation About x-Axis•Rotate the region bounded by y = 4x and y = x2 about the x-axis•What are the dimensions needed?radiusheightthicknessradius = yheight = 4yy -thickness = dy16024yV y y dyp� �= � -� �� ��Rotation About Non-coordinate Axis•Possible to rotate a region around any line•Rely on the basic concept behind the shell methodx = af(x)g(x)2sV radius height t hi cknessp= �� � �Rotation About Non-coordinate Axis•What is the radius?•What is the height?•What are the limits?•The integral:x = af(x)g(x)a – xf(x) – g(x)x = crc < x < a[ ]( ) ( ) ( )acV a x f x g x dx= - -�Try It Out•Rotate the region bounded by 4 – x2 , x = 0 and, y = 0 about the line x = 2•Determine radius, height, limits4 – x2 4 – x2 r = 2 - xr = 2 - x20�Try It Out•Integral for the volume is2202 (2 ) (4 )V x x dxp= - � -�Assignment•Lesson 6.2B•Page 373•Exercises 41 – 59