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I am new to this website, so I apologize if I am doing something wrong.

In the diagram, there are $4$ spheres in a cyclinder. The only information given is this: The radius of each sphere is $5.3$ cm (diameter is $10.6$ cm) And the height of the cylinder $10.6$ cm

How do I find the total surface area of the cylinder?

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Diagram

Edit: to those who have provided a solution, thank you. I will be deleting this soon.

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  • $\begingroup$ I'm trying to understand what you mean by the 'width of the cylinder,' I interpreted that to mean the diameter of the circular cross-section, is that correct? $\endgroup$ – JohnColtraneisJC Aug 12 '17 at 2:00
  • $\begingroup$ The 'width' refers to the rectangular shape (that wraps around the circular face) $\endgroup$ – user471349 Aug 12 '17 at 2:02
  • $\begingroup$ Are the circles arranged into a square? $\endgroup$ – The Great Duck Aug 12 '17 at 2:21
  • $\begingroup$ I only have the diagram to go by. $\endgroup$ – user471349 Aug 12 '17 at 2:22
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    $\begingroup$ "I will be deleting this soon." If you are going to post a question and then delete it after you get answers don't bother posting in the first place! That is really disrespectful to those that took time answering you and is not acceptable practice here. $\endgroup$ – Winther Aug 12 '17 at 2:28
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If you connect centres of the four spheres you obtain a square, whose diagonal, plus two radius at each end, is the diametre of the cylinder.

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If the spheres centers are forming a square, the side of the square will be $2r$.

Referring a coordinate system by putting the square vertex at $(\pm r,\pm r)$, then the cylinder will contact each sphere at angles $45°$, $135°$, $225°$ and $315°$.

So the contact will happen at distance $R=r\sqrt{2}+r$ from the center of the square, which is the radius of the big cylinder. First term is the half size of the square diagonal, and the second term is the radius again.

Then, the area of the cylinder will be $A=hr(\sqrt{2}+1)=10.6 \cdot 5.3(\sqrt{2}+1)=135.6cm^2$

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  • $\begingroup$ Guess it should be $\sqrt{2}r+r$ instead. $\endgroup$ – Vim Aug 12 '17 at 2:18
  • $\begingroup$ corrected...... $\endgroup$ – Brethlosze Aug 12 '17 at 2:20

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