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anonymous

  • one year ago

Example 4 on http://tutorial.math.lamar.edu/Classes/CalcIII/DIPolarCoords.aspx Why is this correct? How does subtracting the volume under z = 16 from the volume under z = x^2 + y^2 on a specific radius give us the desired volume?

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  1. anonymous
    • one year ago
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    Sorry dude the link doesn't work....

  2. anonymous
    • one year ago
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    Is this Calc 3?

  3. anonymous
    • one year ago
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    Sorry, I fixed it now. Yes it is.

  4. anonymous
    • one year ago
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    Wow... I am sorry I can't help. I just started Calculus - I am only in tenth grade. Sorry.

  5. anonymous
    • one year ago
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    No problem :)

  6. IrishBoy123
    • one year ago
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    you're looking for the volume of the paraboloid itself, ie the volume inside that solid but the normal double integral of the function f, ie\( \iint f(x,y) \ dA \) will give you the volume under the pataboloid, ie the volume from the xy plane upwards

  7. IrishBoy123
    • one year ago
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    handily, though, that volume it does give you can be subtracted from the volume of the cylinder to get the inside volume. and by cylinder i mean the volume under the plane \(z = 16\) within the region \(16 = x^2 + y^2\)

  8. phi
    • one year ago
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    In 2D they are doing this |dw:1440248565912:dw| that is the integral under the curve f(x)

  9. phi
    • one year ago
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    but if we want the area (or in 3D, the volume) of the "inside" |dw:1440248674950:dw|

  10. phi
    • one year ago
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    then we find the area of the enclosing rectangle (cylinder in 3D) and subtract off the area of f(x) |dw:1440248748282:dw|

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