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anonymous

  • one year ago

3D coordinate systems

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  1. anonymous
    • one year ago
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    Find the volume of the solid that lies inside both of the spheres x^2 + y^2 + z^2 + 4x -2y + 4z + 5 = 0 x^2 + y^2 + z^2 = 4 I solved the first equation and got: (x+2)^2 + (y-1)^2 + (z+2)^2 = 4 But idk how to finish it

  2. anonymous
    • one year ago
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    You're looking for the volume of the intersection (the sort of lemon-shaped region in the graphic).

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  3. anonymous
    • one year ago
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    Do i have to integrate?

  4. anonymous
    • one year ago
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    I would think so. You can find the volume by setting up a triple integral. The trick is finding the proper limits of integration. I'm not sure if there's a non-calculus way to do it.

  5. anonymous
    • one year ago
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    What if I rotate it across the xy plane? with disks

  6. anonymous
    • one year ago
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    I'm not entirely sure what you mean, but if I'm following it sounds like you mean trying to rearrange one of the spheres so that one is directly above the other. That would certainly make the integral setup easier. Like this, but in 3 dimensions: |dw:1433108834455:dw| You could then theoretically find the volume using the disk/shell method, sure.

  7. anonymous
    • one year ago
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    I just remember my book proved the volume of a sphere this way: |dw:1433108968408:dw|

  8. anonymous
    • one year ago
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    So something like that

  9. anonymous
    • one year ago
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    |dw:1433109006413:dw|

  10. anonymous
    • one year ago
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    Wouldn't it be that area

  11. anonymous
    • one year ago
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    Yes, though I think you should arrange the spheres the other way as in my drawing. This way you only have to set up one integral as opposed to two.

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