Maximum of a triple integral

  • Thread starter phrygian
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  • #1
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Homework Statement



Find the region E for which the triple integral:

(triple integral over E) (1 - x^2 -2y^2 -3z^2) dV is a maximum.

Homework Equations





The Attempt at a Solution



I remember in earlier math courses finding the derivative of a single variable integral, does this problem involve finding the derivative of a triple integral and setting it equal to zero to find the maximum? If so, how would you do that?
 

Answers and Replies

  • #2
tiny-tim
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Hi phrygian! :smile:

You're making this too complicated …

any region in which the integrand is positive will increase the integral, and any region in which the integrand is negative will decrease it …

soooo … ? :wink:
 
  • #3
HallsofIvy
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In other words, where is [itex]1 - x^2 -2y^2 -3z^2\ge 0[/itex]?
 
  • #4
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So the region is when z = sqrt( (-2y^2 - x^2)/3 )? Is this a complete answer how do you describe the region?
 
  • #5
tiny-tim
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Hi phrygian! :smile:

(have a square-root: √ and try using the X2 tag just above the Reply box :wink:)
So the region is when z = sqrt( (-2y^2 - x^2)/3 )? Is this a complete answer how do you describe the region?

erm :redface: … you can't have √ of a negative nnumber, can you? :wink:

Try again. :smile:
 

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