Calculating Average Height of a Constricted Hemisphere

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The discussion focuses on calculating the average height of a constrained hemisphere defined by the equation z=sqrt(a^2-x^2-y^2) within the cone x^2+y^2<=a^2. Participants suggest converting Cartesian coordinates to cylindrical coordinates to simplify the integration process. There is a specific emphasis on using the formula for average height, which involves a double integral over the defined region. Challenges arise in integrating the function r*sqrt(a^2-r^2), with suggestions for using trigonometric substitution. The clarification that the integration is over a hemisphere rather than a cone is also highlighted.
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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