Ch 7 c volumes

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Ap Calculus VOLUMES

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Ch 7 c volumes

  1. 1. Volume Application of Integrals
  2. 2. Volume by Slicing Use when given an object where you know the shape of the base and where perpendicular cross sections are all the same, regular, planar geometric shape.
  3. 3. Volume of a Solid … of known integrable cross section - area A(x) – from a to b, defined as   b a V A x dx 
  4. 4. Procedure: volume by slicing o sketch the solid and a typical cross section o find a formula for the area, A(x), of the cross section o find limits of integration o integrate A(x) to get volume
  5. 5. Example Find the volume of a solid whose base is the circle and where cross sections perpendicular to the x-axis are all squares whose sides lie on the base of the circle. 2 2 4x y 
  6. 6. First, find the length of a side of the square the distance from the curve to the x-axis is half the length of the side of the square … solve for y 2 2 2 2 2 4 4 4 x y y x y x       length of a side is : 2 2 4 x
  7. 7.     2 2 2 2 2 4 4 4 16 4 Area x x x         2 2 2 16 4Volume x dx    128 42.667 3 Answer  
  8. 8. Example A solid has a circular base. Find the volume of the solid if every plane cross section is an equilateral triangle, perpendicular to the x-axis, given the equation of the circle to be … 2 2 9x y 
  9. 9. Solution 2 2 9s x  2 27 3h x      2 21 2 2 9 3 9A x x x    length of a side: height of triangle … found using Pythagorean Theorem: Area: Volume:   3 2 3 3 9 36 3 62.354V x dx     
  10. 10. Volume of a Solid of Revolution A solid of revolution may be generated by revolving the area under the graph of a continuous, non-negative function y = f(x) from a to b, about the x-axis Common Examples cone … generated by revolving the area under a line that passes through the origin, from 0 to k, about the x-axis
  11. 11. cylinder … generated by revolving the area under a horizontal line of positive height from a to b, about the x-axis
  12. 12. Example Look at the region between the curve and the x-axis, from x = 0 to x = 2, and revolve it about the x-axis y x
  13. 13. If we slice the resulting solid perpendicular to the x-axis, each cross section is a circle, or disk The radius of the circle is the distance from the curve to the axis of rotation Area of a cross section (circle)     2 2 / radiuscircleA r x A x xx      
  14. 14. The thickness of each disk is infinitely small, so it is represented as dx. By adding the area of all these circles, from x=0 to x=2, we would get the volume of the solid.   2 0 21 2 ... 2 0 2 V A x dx V x dx x          
  15. 15. Disk Method 2 r radius is always perpendicular to the rotational axis rotational solid with no hollow parts will always have a circular cross section, with area use as long as there is NO hollow space Formula:   2 b a V f x dx   
  16. 16. Example: Disk Method Rotate the area bounded by and the x-axis, about the x-axis; then calculate the volume   2 4f x x   Calculus Applet Volumes of Revolution
  17. 17. x circular cross section2 radius 4x 
  18. 18. Solution    2 4 0 2 2 0 2 2 x x x x x          √ Find limits of integration √ Find radius of cross section 2 4r x  
  19. 19. √ Volume   2 22 2 4V x dx     512 107.233 15   
  20. 20. Example: Disk Method Find the volume generated by rotating the area in the first quadrant bounded by the curve, the y-axis, and the line y = 9, around the y-axis.   2 curve: f x x Use y as variable of integration Calc Applets Volumes of Revolution
  21. 21. circular cross section radius y
  22. 22. Solution x y      9 2 0 V y dy √ limits from y = 0 to y = 9 √ radius √ Answer   81 2
  23. 23. Washer Method used when solid has hollow parts radius of rotation – perpendicular to the axis of rotation two radii - outer and inner Formula R » outer radius r » inner radius            2 2 b a V R x r x dx
  24. 24. Example: Washer Method Find the volume of the solid when is rotated about the line y = -2, from x=0 to x=2  2 y x
  25. 25. outer radius R inner radius r
  26. 26. Solution Find outer and inner radii outer - curve furthest away from the axis of rotation; subtract from this, the axis of rotation      2 2 2 2 x x
  27. 27. inner - inside curve, forms the hollow wall of the solid        0 2 0 2 2 r            2 2 22 0 2 2V x dx
  28. 28.         2 4 2 0 4 4 4x x dx        5 3 32 3241 5 3 5 3 256 53.617 15 x x
  29. 29. Example: Washer Method Find the volume of the solid of revolution generated by revolving about the y-axis, the area enclosed by the graphs of   2 2 andy x y x
  30. 30. outer radius R inner radius r
  31. 31. Solution    2 2 2 2 0 x x x x √ solve for limits           ordered pa 2 0 0 an irs: 0,0 d and 2, 2 4 x x x x
  32. 32. √ to revolve about the y-axis, need to solve each equation for x   1 and 2 x y x y √ radii - in relation to the y-axis, the outer radius is and the inner radius is x y  1 2 x y
  33. 33.              4 2 21 2 0 8 3 V y y dy
  34. 34. Example: Washer Method Consider the area captured between the graphs of     2 2 1 and 1y x x y What volume is generated if this area is rotated about the x-axis ?
  35. 35. √ limits        2 2 2 1 1 2 0 x x x x      0 2 0 2x x x x
  36. 36. √ radii     2 2 1 1R x x r √ answer              2 2 22 0 2 1 1V x x dx    56 11.729 15
  37. 37. Example: Washer Method Find the volume of the solid generated when the region bounded by   ln and 2y x y x is rotated about the line y = -3
  38. 38. √ limits Only way is to obtain from graphing calculator …    3.146, 1.146 .158, 1.841 √ radii                ln 3 ln 3 2 3 1 R x x r x x
  39. 39. √ Answer            3.146 2 2 .158 ln 3 1V x x dx  34.198V

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