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Ap Calculus VOLUMES

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- 1. Volume Application of Integrals
- 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. 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. 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. 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. 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. 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. 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. 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. 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. cylinder … generated by revolving the area under a horizontal line of positive height from a to b, about the x-axis
- 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. 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. 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. 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. 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. x circular cross section2 radius 4x
- 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. √ Volume 2 22 2 4V x dx 512 107.233 15
- 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. circular cross section radius y
- 22. Solution x y 9 2 0 V y dy √ limits from y = 0 to y = 9 √ radius √ Answer 81 2
- 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. 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. outer radius R inner radius r
- 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. 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. 2 4 2 0 4 4 4x x dx 5 3 32 3241 5 3 5 3 256 53.617 15 x x
- 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. outer radius R inner radius r
- 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. √ 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. 4 2 21 2 0 8 3 V y y dy
- 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. √ limits 2 2 2 1 1 2 0 x x x x 0 2 0 2x x x x
- 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. 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. √ 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. √ Answer 3.146 2 2 .158 ln 3 1V x x dx 34.198V

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