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anonymous

  • one year ago

I keep getting 0 as the volume? Giving a medal to whoever can explain it to me correctly! Find the volume of the solid whose base is bounded by the lines F(x) = 1-x/2 , g(x) = -1+x/2 , and x=0. The cross sections perpendicular to the x-axis are equilateral triangles.

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  1. jim_thompson5910
    • one year ago
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    f(x) is this? \[\Large f(x) = \frac{1-x}{2}\] or this? \[\Large f(x) = 1-\frac{x}{2}\]

  2. anonymous
    • one year ago
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    It's \[1-x/2\] and \[-1+x/2\]

  3. jim_thompson5910
    • one year ago
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    so those 1s aren't part of the fraction?

  4. jim_thompson5910
    • one year ago
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    |dw:1444623238793:dw|

  5. jim_thompson5910
    • one year ago
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    |dw:1444623252668:dw|

  6. jim_thompson5910
    • one year ago
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    the base of the solid is this triangle shaded |dw:1444623339028:dw|

  7. jim_thompson5910
    • one year ago
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    |dw:1444623358917:dw|

  8. anonymous
    • one year ago
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    x=2 , the upper limit

  9. jim_thompson5910
    • one year ago
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    yes x = 0 is the lower limit

  10. jim_thompson5910
    • one year ago
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    |dw:1444623523502:dw|

  11. jim_thompson5910
    • one year ago
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    |dw:1444623536918:dw|

  12. anonymous
    • one year ago
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    The length of the black line?

  13. jim_thompson5910
    • one year ago
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    yeah

  14. jim_thompson5910
    • one year ago
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    sorry I should have put it on the other drawing

  15. jim_thompson5910
    • one year ago
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    |dw:1444623622402:dw|

  16. anonymous
    • one year ago
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    Sorry, I don't really know where the line meets the x axis. Is the line at x=1 or x=2?

  17. jim_thompson5910
    • one year ago
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    pick any point on the line sloping downward what is the y coordinate of this point?

  18. jim_thompson5910
    • one year ago
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    what is the y coordinate of this point? in terms of x

  19. jim_thompson5910
    • one year ago
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    that y coordinate would be 1 - (x/2), right?

  20. anonymous
    • one year ago
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    Yeah, the equation for y in terms of x is 1/(x/2). If the 1-x/2 and -1+x/2 weren't the same when squared, wouldn't the equation just be \[\sqrt3/4\int\limits_{0}^{2}(1-x/2)^2-(-1+x/2)^2\] ?

  21. jim_thompson5910
    • one year ago
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    not quite

  22. jim_thompson5910
    • one year ago
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    |dw:1444624174284:dw|

  23. jim_thompson5910
    • one year ago
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    subtract the two (top - bottom) to find the distance [1-(x/2)] - [-1+(x/2)] 1-(x/2) + 1-(x/2) 2-2*(x/2) 2-x so no matter what x value you pick (from x = 0 to x = 3), the distance is 2-x |dw:1444624259423:dw|

  24. anonymous
    • one year ago
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    Oh, I see. Then if 2-x is the distance, what is my next step?

  25. jim_thompson5910
    • one year ago
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    imagine the shaded region in the drawing |dw:1444624423818:dw| forms the floor of this 3D figure

  26. jim_thompson5910
    • one year ago
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    if we look at it from a different angle, we will see |dw:1444624456882:dw|

  27. jim_thompson5910
    • one year ago
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    imagine it from a different perspective

  28. jim_thompson5910
    • one year ago
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    lay out that line that is 2-x units long |dw:1444624481399:dw|

  29. jim_thompson5910
    • one year ago
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    then form an equilateral triangle |dw:1444624510896:dw| each side of the equilateral triangle is 2-x units long

  30. anonymous
    • one year ago
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    Okay..

  31. jim_thompson5910
    • one year ago
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    hopefully you're able to picture what is going on in 3D

  32. jim_thompson5910
    • one year ago
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    what is the area of an equilateral triangle (in general)?

  33. anonymous
    • one year ago
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    \[\sqrt3/4(any side)^2\]

  34. jim_thompson5910
    • one year ago
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    yep

  35. jim_thompson5910
    • one year ago
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    So in this case, \[\Large A = \frac{\sqrt{3}}{4}s^2\] \[\Large A = \frac{\sqrt{3}}{4}(2-x)^2\]

  36. jim_thompson5910
    • one year ago
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    You need to compute \[\Large \int_{0}^{3}Adx\] \[\Large \int_{0}^{3}\frac{\sqrt{3}}{4}(2-x)^2dx\] \[\Large \frac{\sqrt{3}}{4}\int_{0}^{3}(2-x)^2dx\] \[\Large \frac{\sqrt{3}}{4}\int_{0}^{3}(4-4x+x^2)dx\] I'll let you finish up

  37. anonymous
    • one year ago
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    Oh, I see. That actually does make sense. But wouldn't the upper limit be 2 instead of 3?

  38. jim_thompson5910
    • one year ago
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    oh right, not sure how 3 snuck in there like that. so yeah it should be \[\Large \frac{\sqrt{3}}{4}\int_{0}^{2}(4-4x+x^2)dx\]

  39. anonymous
    • one year ago
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    \[(2\sqrt3)/3\] is my answer. Thanks!

  40. jim_thompson5910
    • one year ago
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    correct, you're welcome

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