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anonymous

  • 5 years ago

any body can derive the formula for a surface area and volume of cone through integration(calculus)??????????

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  1. amistre64
    • 5 years ago
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    its been done already :)

  2. anonymous
    • 5 years ago
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    hahaha

  3. amistre64
    • 5 years ago
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    the typical radius is what.... is what, the slope of the linear between radius and height?

  4. amistre64
    • 5 years ago
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    you are adding up all the circumferences of the circles from 0 to h

  5. Yuki
    • 5 years ago
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    volume is easy

  6. Yuki
    • 5 years ago
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    the base of a cone is shaped as a circle

  7. amistre64
    • 5 years ago
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    [S] 2pi [f(x)] dx ...right?

  8. anonymous
    • 5 years ago
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    i tried integrating it bt i guess well have 2 use the volume of a frustum

  9. Yuki
    • 5 years ago
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    so the base area is \[\pi r^2\]

  10. Yuki
    • 5 years ago
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    if you use the disc method to integrate it into a cone, first we need to find the volume of each disc

  11. amistre64
    • 5 years ago
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    do we include the area of the base into surface area? or is it an open base?

  12. anonymous
    • 5 years ago
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    the disc method dint wrk out

  13. Yuki
    • 5 years ago
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    if dx is the width of each disc, then the volume would be \[\pi r^2dx\]

  14. anonymous
    • 5 years ago
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    i tried this

  15. amistre64
    • 5 years ago
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    r = [f(x)] right?

  16. Yuki
    • 5 years ago
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    now since the radius of a cone changes as the position of the disc changes, we need to find r in terms of x

  17. amistre64
    • 5 years ago
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    f(x) = h = linear equation from point (0,r) to (r,h)

  18. amistre64
    • 5 years ago
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    or... (0,x) to (x,y) :)

  19. amistre64
    • 5 years ago
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    ack...(x,0.. my mistake)

  20. anonymous
    • 5 years ago
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    dV= πr2^2dx

  21. amistre64
    • 5 years ago
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    him; deriving to get volume?

  22. amistre64
    • 5 years ago
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    integrate

  23. Yuki
    • 5 years ago
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    if a cone has radius r and height h, the radius of the disc will be \[({-r \over h}x +r)\]

  24. anonymous
    • 5 years ago
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    r=xtant

  25. amistre64
    • 5 years ago
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    do we have a height and a radius?

  26. Yuki
    • 5 years ago
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    so the volume of the disc is \[\pi ({-r \over h}x+r)^2dx\]

  27. Yuki
    • 5 years ago
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    now we will find the limits of integration

  28. Yuki
    • 5 years ago
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    this one is easy, the height is h, so the limit of integration is from x=0 to x=h thus the integral we are looking for to find the volume is \[\int\limits_{0}^{h} \pi ({-r \over h}x+r)^2dx\]

  29. Yuki
    • 5 years ago
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    when you integrate it you will get \[{1 \over 3}\pi r^2\]

  30. anonymous
    • 5 years ago
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    dV=πr^2dh r=Rh/H dh=Hdr/R dV=πr^2Hdr/R V= πR^3 x H/3R

  31. Yuki
    • 5 years ago
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    did it help at all ?

  32. anonymous
    • 5 years ago
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    yeah i think this takes care f d volume

  33. Yuki
    • 5 years ago
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    now let's work on the surface area

  34. anonymous
    • 5 years ago
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    dS=2πrdl l=rL/R dl= Ldr/R S= 2Lπrdr/R πR^2 L/R = πRL

  35. anonymous
    • 5 years ago
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    DONE!!!!!!!!!!

  36. Yuki
    • 5 years ago
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    the same idea is applied to the problem, but now instead of a volume, we are looking for an area. so instead of a disc, we will consider the lateral area of the disc, which can be found by \[2\pi r *dx\]

  37. anonymous
    • 5 years ago
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    thanks a lot fr the solution

  38. Yuki
    • 5 years ago
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    again, the radius of a cone changes, so we need to find the radius in terms of x. But we already found it, which is again \[{-r \over h}x+r\]

  39. Yuki
    • 5 years ago
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    again, the radius of a cone changes, so we need to find the radius in terms of x. But we already found it, which is again \[{-r \over h}x+r\]

  40. Yuki
    • 5 years ago
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    so the lateral area of the discs can be found by \[2\pi[ {({-r \over h} )x+r}]dx\]

  41. Yuki
    • 5 years ago
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    now for this one, you have to be careful with the limit of integration

  42. Yuki
    • 5 years ago
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    the length of the slanted height is going to be my limit of integration because we are finding the lateral area of each discs so that all dx's add up to the slanted height.

  43. Yuki
    • 5 years ago
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    so if we could find the slanted height we are good to go

  44. Yuki
    • 5 years ago
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    now you can see that the slanted height makes a right triangle with the height of the cone and the radius of the base

  45. Yuki
    • 5 years ago
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    so if I call L my slanted height, \[L = \sqrt{h^2 + r^2}\]

  46. anonymous
    • 5 years ago
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    S.A = L*2pi*r, where L is the slant length V = A*h, in this case height is dx, r = x so we have; \[\pi\int_0^x x^2 dx = (1/3)\pi x^3 \]

  47. Yuki
    • 5 years ago
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    so the integration we have to do is \[\int\limits_{0}^{L} 2\pi[({-r \over h})x+r]dx\]

  48. Yuki
    • 5 years ago
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    which ends up being \[\pi rL\]

  49. Yuki
    • 5 years ago
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    or \[\pi r \sqrt{h^2+r^2}\]

  50. Yuki
    • 5 years ago
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    I hope that helps :)

  51. Yuki
    • 5 years ago
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    Give me about 5min to get the picture of why the radius is \[{-r \over h}x+r\]

  52. anonymous
    • 5 years ago
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    dV=πr^2dh r=Rh/H dh=Hdr/R dV=πr^2Hdr/R V= πR^3 x H/3R

  53. anonymous
    • 5 years ago
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    thats the vol

  54. Yuki
    • 5 years ago
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  55. Yuki
    • 5 years ago
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    It would be nice if you could tell me if it helped or not ... :)

  56. anonymous
    • 5 years ago
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    ya it helped thanks

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