Calculating Average Height of a Constricted Hemisphere

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Homework Help Overview

The problem involves calculating the average height of a constricted hemisphere defined by the equation z=sqrt(a^2-x^2-y^2) within the constraints of a cone given by x^2+y^2<=a^2 in the xy plane.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss converting Cartesian coordinates to cylindrical coordinates and polar coordinates to facilitate integration. There are questions regarding the integration process, particularly with the expression r*sqrt(a^2-r^2) and the use of trigonometric substitution.

Discussion Status

Some participants have provided suggestions for approaches, including coordinate transformations and clarifications about the geometric interpretation of the region of integration. Multiple interpretations of the problem are being explored, particularly regarding the shape of the region involved.

Contextual Notes

There is a noted confusion regarding the integration region, with participants clarifying that the integration is over a hemisphere rather than a cone.

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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