Calculating Average Height of a Constricted Hemisphere

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SUMMARY

The discussion focuses on calculating the average height of a constricted hemisphere defined by the equation z=sqrt(a^2-x^2-y^2) within the region constrained by the cone x^2+y^2<=a^2. The average height is determined using the formula Average Height =(1/area)*double integral of region of [z]drdpheta. Participants suggest converting Cartesian coordinates to cylindrical coordinates to simplify the integration process, specifically integrating r*sqrt(a^2-r^2) using polar coordinates. The key insight is recognizing that the integration is performed over a hemisphere rather than a cone.

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  • Understanding of double integrals in calculus
  • Familiarity with cylindrical coordinates
  • Knowledge of polar coordinate transformations
  • Experience with trigonometric substitutions in integration
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  • Study the process of converting Cartesian coordinates to cylindrical coordinates
  • Learn about double integrals over polar coordinates
  • Review techniques for trigonometric substitution in calculus
  • Explore applications of average value calculations in multivariable calculus
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Students studying multivariable calculus, particularly those tackling integration problems involving geometric shapes like hemispheres and cones.

jimbo71
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Homework Statement


find the anverage heigh of z=sqrt(a^2-x^2-y^2) constricted by the cone x^2+y^2<=a^2
in the xy plane


Homework Equations


Average Height =(1/area)*double integral of region of [z]drdpheta


The Attempt at a Solution


I really have no idea how to solve this problem can you please point me in the right direction
 
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How about converting your Cartesian functions over to cylindrical coordinates?
 
converted the cartesian equations to polar and used the 1/area*double integral of region R [z]rdrdpheta. I am having much difficulty integrating r*sqrt(a^2-r^2). i tried a trig substitution but don't know how to finish from there. please help me!
 
replace z in that "1/area*double integral of region R [z]rdrdpheta" you wrote by what it is equal to looking at the surface. then you can use polar coordinates.
 
If it's any help, you aren't integrating over a cone, you are integrating over a hemisphere.
 

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