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January 7th, 2018, 07:42 PM   #1
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Anyone can help in this question? Thanks

Question as in attachement, thank you.
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 January 8th, 2018, 02:56 AM #2 Senior Member   Joined: Oct 2009 Posts: 627 Thanks: 190 What is the shape of the figure? What does it look like graphically?
January 8th, 2018, 03:39 AM   #3
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Quote:
 Originally Posted by Micrm@ss What is the shape of the figure? What does it look like graphically?
It doesn’t mentioned the shape, this is the whole question about..

 January 8th, 2018, 04:00 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, 04:57 AM   #5
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Quote:
 Originally Posted by alicia13179 It doesn’t mentioned the shape, this is the whole question about..
Something like $x^2 + y^2 + z^2 = 1$, what is that object? Hint: Pythagoras, distance formula.

If you don't see it immediately, what about $x^2 + y^2 = 1$.

 January 8th, 2018, 11:03 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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 $\int_0^\sqrt{99}\int_0^{2\pi} (\sqrt{100- r^2}- 1) rdrd\theta$. Thanks from Ould Youbba

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