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January 7th, 2018, 06:42 PM  #1 
Newbie Joined: Jan 2018 From: kapit Posts: 2 Thanks: 0  Anyone can help in this question? Thanks
Question as in attachement, thank you.

January 8th, 2018, 01:56 AM  #2 
Senior Member Joined: Oct 2009 Posts: 263 Thanks: 90 
What is the shape of the figure? What does it look like graphically?

January 8th, 2018, 02:39 AM  #3 
Newbie Joined: Jan 2018 From: kapit Posts: 2 Thanks: 0  
January 8th, 2018, 03:00 AM  #4 
Newbie Joined: Jan 2018 From: Tunis, Tunisia Posts: 10 Thanks: 4 Math Focus: Analysis 
Hi ! Did you try to use the cylindrical coordinates or maybe the spherical coordinates ? 
January 8th, 2018, 03:57 AM  #5 
Senior Member Joined: Oct 2009 Posts: 263 Thanks: 90  
January 8th, 2018, 10:03 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,966 Thanks: 807 
The plane $z= 1$ cuts the ball $x^2+ y^2+ z^2\le 100$ where $x^2+ y^2+ 1= 100$ so the disk $x^2+ y^2\le 99$ in the plane z= 1. Some have suggested the use of cylindrical coordinates. We cover that disk by taking r from 0 to $\sqrt{99}$, $\theta$ from 0 to $2\pi$. The height of the sphere above the plane at each r, $\theta$, is $\sqrt{100 x^2 y^2}1= \sqrt{100 r^2}$ and the "differential of area" is $r drd\theta$. So the volume is given by $\int_0^\sqrt{99}\int_0^{2\pi} (\sqrt{100 r^2} 1) rdrd\theta$.


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