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blackstreet23

  • one year ago

What was the calculus formula to determine the volume of a solid? I don't mean solid of revolution. Just regular solid.

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  1. blackstreet23
    • one year ago
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    volumes of cross sections

  2. freckles
    • one year ago
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    Are you sure you aren't looking for: \[\text{ Volume }=\int\limits_a^bA(x) dx \text{ where } A(x) \text{ is the area of the cross section } ?\]

  3. blackstreet23
    • one year ago
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    yes that is it ! thank you !

  4. freckles
    • one year ago
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    that's also the formula used in solids formed by revolution of some curve about some axis

  5. LynFran
    • one year ago
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    u forget to square it..i think @freckles

  6. freckles
    • one year ago
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    you know where we take circles as the cross section

  7. freckles
    • one year ago
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    no A(x) is the area you don't want to square the area

  8. LynFran
    • one year ago
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    o ok

  9. freckles
    • one year ago
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    if the cross section is a circle then A(x)=pi*r^2

  10. freckles
    • one year ago
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    where r is a function of x

  11. freckles
    • one year ago
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    though we could look at things in terms of y too not just x

  12. freckles
    • one year ago
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    you know depending on what suits us better

  13. blackstreet23
    • one year ago
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    but how is that the formula for solids of revolution? I mean the formula for solids of revolution is \[\int\limits_{a}^{b}(\pi*f(x)^2) dx\]

  14. LynFran
    • one year ago
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    i think the f(x) is actually the radius

  15. blackstreet23
    • one year ago
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    f(x) is the function

  16. blackstreet23
    • one year ago
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    i mean the height of the funtion

  17. LynFran
    • one year ago
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    yes which is intern the radius so A=pi*r^2

  18. LynFran
    • one year ago
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    and freckles just use the Area instead...i think

  19. blackstreet23
    • one year ago
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    I guess i am kind of confuse lol

  20. LynFran
    • one year ago
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    ok lets wait on till @freckles come online...

  21. freckles
    • one year ago
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    what's the question?

  22. freckles
    • one year ago
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    area of a circle is what I called A(x)

  23. freckles
    • one year ago
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    or the are of the cross section is A(x)

  24. freckles
    • one year ago
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    if the cross section is a circle then A(x) is pi*r^2 where r is the radius f(x) is the distance from the curve to the x-axis so f(x) is the radius if the axis of rotation is the x-axis

  25. freckles
    • one year ago
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    |dw:1440894543729:dw| Say we want to take this curve on [a,b] and rotate it about the x-axis

  26. freckles
    • one year ago
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    first reflecting the curve about the x-axis gives us: |dw:1440894612771:dw| now that drawing is not that great but that curve is basically suppose to be a mirror for the other curve I drew

  27. freckles
    • one year ago
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    |dw:1440894662165:dw| so we have a cross section that is a circle

  28. freckles
    • one year ago
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    notice the distance from the x-axis to the curve is indeed the radius which is called f(x) in this case

  29. freckles
    • one year ago
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    \[\int\limits_{a}^{b}A(x) dx \\ \text{ so we have } A(x)=\pi \cdot (f(x))^2 \\ \]

  30. LynFran
    • one year ago
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    so i was right about f(x) being the radius?

  31. blackstreet23
    • one year ago
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    ohh i see

  32. freckles
    • one year ago
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    |dw:1440894830121:dw| yes

  33. freckles
    • one year ago
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    but the A(x) is not squared

  34. freckles
    • one year ago
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    A(x) just means area of the cross section

  35. blackstreet23
    • one year ago
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    ohh i see ! I thought f(x) was just the height of the function

  36. blackstreet23
    • one year ago
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    It makes so much sense now :)

  37. freckles
    • one year ago
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    well you can also say |f(x)| is the height of the function

  38. freckles
    • one year ago
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    I put | | around it because f(x) could happen to be negative which just means the height is below the x-axis

  39. LynFran
    • one year ago
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    cool..

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