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anonymous

  • one year ago

The base of a solid is bounded by the curve y = sqrt(x + 1), the x-axis and the line x = 1. The cross sections, taken perpendicular to the x-axis, are squares. Find the volume of the solid 1 2 2.333 None of these

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  1. anonymous
    • one year ago
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    @freckles @ganeshie8 @Hero @nincompoop @Robert136 @satellite73

  2. anonymous
    • one year ago
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    2, I think.

  3. anonymous
    • one year ago
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    dont really want the answer as much as i want to know how to do it :P

  4. triciaal
    • one year ago
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  5. rds8701
    • one year ago
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    PLEASE HELP WILL FAN AND MEDAL According to an article in Runners' World magazine: On average the human body is more than 50 percent water [by weight]. Runners and other endurance athletes average around 60 percent. This equals about 120 soda cans' worth of water in a 160-pound runner! Investigate their calculation. Approximately how many soda cans’ worth of water are in the body of a 160-pound runner? What unprovided information do you need to answer this question?

  6. anonymous
    • one year ago
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    wingspan move man ... wingspan move.

  7. triciaal
    • one year ago
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    I think we need the area under the curve between 0 and 1 times the height

  8. SolomonZelman
    • one year ago
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    Is the region bounded by sqrt(x+1), y=0, and x=1 rotated around some line? Because without that, so far, you only have a 2D object....

  9. triciaal
    • one year ago
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    |dw:1439477813129:dw|

  10. anonymous
    • one year ago
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    its not rotated its |dw:1439477867812:dw| this is what it should loo like

  11. triciaal
    • one year ago
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    when y = 0 then x = -1

  12. anonymous
    • one year ago
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    i don't really see how that's useful

  13. triciaal
    • one year ago
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    did you see the graph?

  14. anonymous
    • one year ago
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    yes i have a ti-84 iv graphed it , i have a table of values, don't see the use to it though

  15. SolomonZelman
    • one year ago
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    Never claimed to be good at math, but to me the question is missing, because as it is the very region you are dealing with is 2D.

  16. anonymous
    • one year ago
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    pretty sure that its a complete question and that the object described is 3d

  17. anonymous
    • one year ago
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    @ParthKohli @OregonDuck any ideas ?

  18. SolomonZelman
    • one year ago
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    It is just bound by some (explicit) functions, and there is no reason for it to be 3D. Is it rotated about some line? You said it is not.

  19. anonymous
    • one year ago
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    it says that the cross sections perpendicular to the x axis are squares, only 3d objects have cross sections

  20. triciaal
    • one year ago
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    so we need the integral between -1 and 1

  21. anonymous
    • one year ago
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    already did that 1.8856

  22. anonymous
    • one year ago
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    @pooja195

  23. triciaal
    • one year ago
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    we have half a parabola that opens right guess none of the above

  24. freckles
    • one year ago
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    |dw:1439482917191:dw|

  25. freckles
    • one year ago
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    |dw:1439483111182:dw| I'm bad about drawing 3d objects but this is what the shape looks like

  26. freckles
    • one year ago
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    sorta

  27. freckles
    • one year ago
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    |dw:1439483183181:dw|

  28. freckles
    • one year ago
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    @jdosio do you see how to setup the integral let me know when you get back on

  29. freckles
    • one year ago
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    in other words you need to find the area of the square

  30. freckles
    • one year ago
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    and use that as the integrand

  31. anonymous
    • one year ago
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    damn now your offline, ill be on for the remainder of the day @freckles tell me when you get back on

  32. anonymous
    • one year ago
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    also thats the only part i have a problem with,(setting up the integral ) i know what the solid looked like and all just not sure how to calculate its volume

  33. freckles
    • one year ago
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    \[V=\int\limits_{-1}^1 A(x) dx \\ \text{ where } A(x)=\text{ area of square }\] |dw:1439494770565:dw|

  34. freckles
    • one year ago
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    \[A(x)=\text{ area of square}=\text{ height } ^2\]

  35. freckles
    • one year ago
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    the integrand is just x+1 that is A(x)=x+1

  36. freckles
    • one year ago
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    \[V=\int\limits_{-1}^1 (x+1) dx\]

  37. anonymous
    • one year ago
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    isint it \[\sqrt{x+1}\]

  38. freckles
    • one year ago
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    that is the height you also have the base is that too the area of a square=base*height or height^2 since base=height

  39. freckles
    • one year ago
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    |dw:1439495278485:dw| if that length here is sqrt(x+1) then the base length of that shape is also sqrt(x+1) since the shape is a square

  40. freckles
    • one year ago
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    |dw:1439495315876:dw|

  41. freckles
    • one year ago
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    \[A(x)=\sqrt{x+1} \cdot \sqrt{x+1}=(x+1)\]

  42. anonymous
    • one year ago
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    !! i get it !!

  43. freckles
    • one year ago
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    notice the lower limit is x=-1 and the upper limit is x=1

  44. freckles
    • one year ago
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    \[V=\int\limits _{-1}^1 A(x) dx \\ A(x)=\sqrt{x+1} \cdot \sqrt{x+1}=x+1 \\ V=\int\limits_{-1}^{1} (x+1) dx \] V represents the volume by the way

  45. anonymous
    • one year ago
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    okay im getting 2

  46. freckles
    • one year ago
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    sounds great that is what @IrishBoy123 got using double integrals

  47. anonymous
    • one year ago
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    thanks for your help freckles , cant thank you enough

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