Infinite Cake: find height, volume, surface area

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ecxtimmy

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I have a project problem for my Calc 3 class, its as follows

An infinte cake is created as follows. The bast of the cake is cylindrical. It is four inches tall and 12 inches in diameter. There are infintely many cylindrical laers in the cake, and each layer is half as tall and has (2/3) the diameter of the previous layer.

1) How tall is the cake

2) What is the total volume of the cake?

3) What is the surface are of the cake? This part is the trickiest, because you should compute exactly the amount of cake that would beed ot be frosted. Don't frost parts of the cake that are covered by other layers. Also don't frost the bottom of the base.

If any of the answers above is "infinity" , then clearly explain why this is true.
 
Re: Infinite Cake

ecxtimmy said:
I have a project problem for my Calc 3 class, its as follows

An infinte cake is created as follows. The bast of the cake is cylindrical. It is four inches tall and 12 inches in diameter. There are infintely many cylindrical laers in the cake, and each layer is half as tall and has (2/3) the diameter of the previous layer.

1) How tall is the cake

h = 4 + 4(1/2) + 4(1/2)[sup:2mv6zqyh]2[/sup:2mv6zqyh] + 4(1/2)[sup:2mv6zqyh]3[/sup:2mv6zqyh] + ...

2) What is the total volume of the cake?

V = pi*6[sup:2mv6zqyh]2[/sup:2mv6zqyh]*4 + pi*[6(2/3)][sup:2mv6zqyh]2[/sup:2mv6zqyh]*4(1/2) + pi*[6(2/3)[sup:2mv6zqyh]2[/sup:2mv6zqyh]][sup:2mv6zqyh]2[/sup:2mv6zqyh]*4(1/2)[sup:2mv6zqyh]2[/sup:2mv6zqyh] + pi*[6(2/3)[sup:2mv6zqyh]3[/sup:2mv6zqyh]][sup:2mv6zqyh]2[/sup:2mv6zqyh]*4(1/2)[sup:2mv6zqyh]3[/sup:2mv6zqyh] + ...

3) What is the surface are of the cake? This part is the trickiest, because you should compute exactly the amount of cake that would beed ot be frosted. Don't frost parts of the cake that are covered by other layers. Also don't frost the bottom of the base.

total vertical surface area = 2pi*6*4 + 2pi*6(2/3)*4(1/2) + 2pi*6(2/3)[sup:2mv6zqyh]2[/sup:2mv6zqyh]*4(1/2)[sup:2mv6zqyh]2[/sup:2mv6zqyh] + 2pi*6(2/3)[sup:2mv6zqyh]3[/sup:2mv6zqyh]*4(1/2)[sup:2mv6zqyh]3[/sup:2mv6zqyh] + ...
total horizontal surface area = pi*6[sup:2mv6zqyh]2[/sup:2mv6zqyh]

If any of the answers above is "infinity" , then clearly explain why this is true.
Click to expand...
 
Re: Infinite Cake

Using geometric series laws im getting that the hieght reaches a limit of 8. What about the other 2. It would seem that if the height is finite that the surface area is also finite but im not sure if the volume is infinte.

also applying geometric properties to the volume im getting that it converges to (1296 * pi)/6

Im still alittle bit confused about the surface area. The bottom doesn't need to be frosted but the sides and top do, and since only the final layer needs to be frosted it would be the nth term and n goes to infinity
 
Re: Infinite Cake

Here's a clue ... the other "two" are also infinite geometric series with a common ratio < 1 ... find the common ratio for each and apply the expression for the sum of an infinite geometric series.
 
Your volume is slighly off.

Factor out the \(\displaystyle 6^{2}\cdot{4}\cdot{\pi}=144{\pi}\)

Combine it all into one sum.

\(\displaystyle 144{\pi}\sum_{k=0}^{\infty}\left(\frac{2}{3}\right)^{2k}\cdot\sum_{k=0}^{\infty}\left(\frac{1}{2}\right)^{k}\)

\(\displaystyle =144{\pi}\sum_{k=0}^{\infty}\left(\frac{2}{9}\right)^{k}=\frac{1296\pi}{7}\)

For the surface area. You are not counting the top or bottom, just the sides of the 'cylinders'.

We use \(\displaystyle 2{\pi}rh\)

So, simplifying we get the sum:

\(\displaystyle \text{surface area=}48{\pi}\sum_{k=0}^{\infty}\frac{1}{3^{k}}=72{\pi}\)
 
Hey Gal, im assuming that your K = n-1 which would be excatly what i have. cause then it would be (144*pi) / (1 - (2/9)) which gives (1296*pi) / 7
 
You may have it set up a little different, but that's what I get also. :)
 
Ya because u set it up with K starting from 0, were as I use N-1 but start from 1, so they are equivalent.

Thanks
 
I already answered this question ... look back at my first post in the this thread.
 
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