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 #32
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Saturday 7/6/14

http://web2.0calc.com/questions/attn-registered-users#r8

Attitude - reminder and comedy.

http://web2.0calc.com/questions/need-help-asap_4#rr1

This one is still annoying me.

 http://web2.0calc.com/questions/domain-and-range-of-80n-275

 Jibberish.

http://web2.0calc.com/questions/what-is-2-plus-5-log-times-10000-4956-npr

 Binomial expansion

http://web2.0calc.com/questions/use-the-binomial-theorem-to-expand-the-binomial-x-7y-3-and-simplyfy

I thought some people might be interested in how Chris and i answered this simple question.  We could have answered better but I think it is ok.

http://web2.0calc.com/questions/is-0-5-larger-than-0-5

I liked this question - It needs to be checked.

http://web2.0calc.com/questions/from-cube-to-tetraedre

 Square root in quehttp://web2.0calc.com/questions/from-cube-to-tetraedrestion verses Square root introduced in the working.

I have added the address of this one to the Sticky Thread called "Information worth keeping or Developing"

 http://web2.0calc.com/questions/answer-of-square-root-4-must-be-plus-minus-2

 

Melody....could you please highlight this one on your "Daily Wrap??"

I thought it was excellent...... 
http://web2.0calc.com/questions/what-is-the-surface-are-of-a-cylinder-with-r-8-h-4#r2 
A very nice explanation by NinjaDevo!!!......I'm gonna "steal" it for future reference...!!! 

Jun 6, 2014
 #5
avatar+2354 
0

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Jun 6, 2014
 #3
avatar+118723 
+5

I love it Chris!

Jun 6, 2014
 #1
avatar+2354 
+6

This image might help;

Now the surface area basically consists of three parts, the top, the bottom and what i'll call the wrapper (I dont know the correct English word). Remember that you could calculate the surface of a circle with the formula  $$\pi r^2$$ where r is the radius.

Therefore if we divide the volume by the height, we'll know the surface of the top of the cylinder.

The height (What is described in your case as length, but this is just a matter of how you hold the cylinder) is 25m

Therefore the surface of the top circle is $$\frac{400m^3}{25m} = 16m^2$$

To find the 'wrapper' we first need to find the radius of the circle.

We know the surface of a circle is $$\pi r^2$$ therefore 

$$\pi r^2 = 16m^2 \Rightarrow r = \sqrt{\frac{16m^2}{\pi}} \approx 5.09m$$

 

Now to calculate the 'wrapper' think of this area as if I wanted to wrap a paper around this part, how much paper would I need?

If I'd wrap a perfectly matching paper around the side of the cylinder it would have the length of the circumference of the cylinder and the height would be the same as that of the cylinder.

 

Therefore, we want to find the circumference (let's call it C) of the circle.

The formula for this is

$$C = 2 \pi r$$

And if we want to know the entire surface of the wrapper (let's call this W) we need to multiply this by the height

$$W = 2 \pi r h$$

Therefore

$$W = 2 \pi \sqrt{\frac{16m^2}{\pi}}25m$$

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\sqrt{{\frac{{\mathtt{16}}}{{\mathtt{\pi}}}}}}{\mathtt{\,\times\,}}{\mathtt{25}} = {\mathtt{354.490\: \!770\: \!181\: \!103\: \!182\: \!2}}$$

Therefore the surface of the wrapper is approx. 354.5m^2

the surface of the top is 16m^2

the surface of the bottom is the same as the top and therefore also 16m^2

And hence

The surface area of the cylinder is

$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{16}}{\mathtt{\,\small\textbf+\,}}{\mathtt{354.49}} = {\mathtt{386.49}}$$

386.49m^2

Jun 6, 2014

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