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anonymous

  • one year ago

The base of a solid is the circle x^2 + y^2 = 9. Cross sections of the solid perpendicular to the x-axis are equilateral triangles. What is the volume, in cubic units, of the solid?

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  1. anonymous
    • one year ago
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    @jim_thompson5910 @zepdrix

  2. anonymous
    • one year ago
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    Would the radius not be 3?

  3. anonymous
    • one year ago
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    https://i.gyazo.com/f3d88835c49eb9270050edba54fcec2d.png

  4. jim_thompson5910
    • one year ago
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    this isn't a volume of revolution

  5. anonymous
    • one year ago
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    Hmmmm

  6. jim_thompson5910
    • one year ago
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    it's kinda like this https://www.youtube.com/watch?v=tmotNcRHr9Q

  7. jim_thompson5910
    • one year ago
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    instead of gluing a bunch of disks together, you're gluing a bunch of equilateral triangles together

  8. jim_thompson5910
    • one year ago
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    |dw:1444448060321:dw|

  9. anonymous
    • one year ago
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    So it would look like a cone?

  10. anonymous
    • one year ago
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    Since the base is a circle

  11. jim_thompson5910
    • one year ago
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    kinda, more like the sydney opera house https://upload.wikimedia.org/wikipedia/commons/7/75/Sydney_Opera_House,_botanic_gardens_1.jpg

  12. jim_thompson5910
    • one year ago
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    |dw:1444448113459:dw|

  13. jim_thompson5910
    • one year ago
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    |dw:1444448131880:dw|

  14. jim_thompson5910
    • one year ago
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    hint: solve x^2 + y^2 = 9 for y

  15. anonymous
    • one year ago
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    \[\sqrt{9-x^2}\]

  16. jim_thompson5910
    • one year ago
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    yep

  17. jim_thompson5910
    • one year ago
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    by symmetry, the distance between point A and B is \(\Large 2\sqrt{9-x^2}\) |dw:1444448401555:dw|

  18. anonymous
    • one year ago
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    ok, what am I supposed to do with that?

  19. jim_thompson5910
    • one year ago
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    ok imagine that circle is the floor of this weird 3D object

  20. jim_thompson5910
    • one year ago
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    |dw:1444448545776:dw|

  21. jim_thompson5910
    • one year ago
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    we're going to build a triangular roof like so |dw:1444448565044:dw|

  22. jim_thompson5910
    • one year ago
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    each triangle is an equilateral triangle

  23. jim_thompson5910
    • one year ago
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    so what we do really is integrate a bunch of these triangles together to form the solid so we do \[\Large \int_{-3}^{3}Adx\] where A is the area of any given equilateral triangle

  24. jim_thompson5910
    • one year ago
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    area of equilateral triangle with side length s \[\Large A = \frac{\sqrt{3}}{4}*s^2\]

  25. jim_thompson5910
    • one year ago
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    \[\Large A = \frac{\sqrt{3}}{4}*s^2\] \[\Large A = \frac{\sqrt{3}}{4}*\left(2\sqrt{9-x^2}\right)^2\] \[\Large A = \frac{\sqrt{3}}{4}*\left(4(9-x^2)\right)\] \[\Large A = \sqrt{3}(9-x^2)\]

  26. anonymous
    • one year ago
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    Ok, now I'm following you, but what would I enter for X?

  27. jim_thompson5910
    • one year ago
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    you'll be integrating with respect to x

  28. jim_thompson5910
    • one year ago
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    \[\Large \int_{-3}^{3}Adx = \int_{-3}^{3}\left[\sqrt{3}(9-x^2)\right]dx = ??\]

  29. anonymous
    • one year ago
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    36sqrt(3)

  30. jim_thompson5910
    • one year ago
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    very good

  31. anonymous
    • one year ago
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    I have some other questions that I solved, do you mind checking them for me?

  32. jim_thompson5910
    • one year ago
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    alright

  33. anonymous
    • one year ago
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    Or should I post it separately?

  34. jim_thompson5910
    • one year ago
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    separately to avoid lag and clutter

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