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anonymous
 one year ago
3D coordinate systems
anonymous
 one year ago
3D coordinate systems

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find 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 + (y1)^2 + (z+2)^2 = 4 But idk how to finish it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You're looking for the volume of the intersection (the sort of lemonshaped region in the graphic).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do i have to integrate?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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 noncalculus way to do it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What if I rotate it across the xy plane? with disks

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just remember my book proved the volume of a sphere this way: dw:1433108968408:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So something like that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433109006413:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wouldn't it be that area

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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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