anonymous
  • anonymous
Through the center on two different sized wooden balls there is a hole drilled in different size such that the two " rings " which remains of the balls have the same height. Which ring is heaviest ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Smartyprincess
  • Smartyprincess
The one which the most weight should be the one with the most area. The area is the volume of the sphere, minus the area of the cylinder who represent the hole. But how to compare the two cases?
Smartyprincess
  • Smartyprincess
:)
MrNood
  • MrNood
@Smartyprincess it is VOLUME not Area you need to consider

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Smartyprincess
  • Smartyprincess
oh, sorry
Smartyprincess
  • Smartyprincess
I know that that is showing if the area is there subtract that and get volume which shows which one is bigger
anonymous
  • anonymous
|dw:1449483095910:dw|
anonymous
  • anonymous
h is given by: \[\sqrt{r^{2}_{1} - x ^{2}} =b _{1} \] \[x ^{2} = r ^{2}_{1}-b ^{2}_{1}\rightarrow x= \sqrt{r ^{2}_{1}-b ^{2}_{1}} \leftarrow \rightarrow h= \sqrt{r ^{2}_{1}-b ^{2}_{1}} \]
anonymous
  • anonymous
and the volym after the drilling is \[\pi \int\limits_{0}^{h}r ^{2}_{1}-b ^{2}_{1} dx -\pi \int\limits_{0}^{h}b^{2}_{1} dx \rightarrow \pi \int\limits_{0}^{h} r ^{2}_{1}-x ^{2}-b ^{2}_{1} dx\]
anonymous
  • anonymous
volume*
anonymous
  • anonymous
so the little ball has \[\frac{ 2\pi(r ^{2}_{1}-b ^{2}_{1})^{\frac{ 3 }{ 2 }} }{ 3 } Volume.Units\]
anonymous
  • anonymous
and the big ball, will have \[\frac{ 2\pi(r ^{2}_{2}-b ^{2}_{2})^{\frac{ 3 }{ 2 }} }{ 3}Volume.Units\] since h is the same? so which one weights the most?
anonymous
  • anonymous
@Michele_Laino

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