How do I integrate a hemisphere using spherical coordinates?

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SUMMARY

This discussion focuses on integrating a hemisphere using spherical coordinates, specifically for a hemisphere with a radius of 10 and an area of 628.3. The user initially approximated the integral using discrete summation, yielding a value of 352.7, but seeks the exact solution through integration of the function y = 100x/(√20x)³ multiplied by the area of a circle. The user notes a 16% difference between the approximate and exact solutions and questions its significance. The integration process is suggested to be performed using triple integrals in spherical coordinates.

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Integrating a hemisphere

Homework Statement


I am self-teaching calculus.
I'd like to integrate a hemisphere G (http://en.wikipedia.org/wiki/Spherical_cap), where the points of each circle have a given property (k=100x/(√20x)3
the radius of G r = 10, A = 628.3, h = x
the top point is A (x=0), a = √20x-x2, and the line AP joining a to a point P is √x2+20x-x2 = √20 x

I learned to find an approximation by discrete summation, finding the value of y for every slice of G of height 1, (whose area is 20 pi = 62.8) (http://www.wolframalpha.com/input/?i=(sum+[+100x/sqrt(20x)^3,{x+,1,10}])+*20pi) y =352.7

Now, I'd like to find the exact solution, integrating the product of y by the area of a single circle
h , I suppose x must be = h as dx must be infinitesimal, is that right?
http://www.wolframalpha.com/input/?i=integrate+y+=+100+*x/sqrt(20*x)^3+*20pi+x+from+1+to+10) y= 304
There is a 16% difference, is it too much or everything is OK?

Thanks
 
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