Volumes of Revolution Word Problem

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  • #1
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Homework Statement



Assume that the Earth is a sphere with circumference of 24,900 miles.
a. Find the volume of the Earth north of latitude 45 degrees. (hint: integrate with respect to y)
b. Find the volume of the Earth between the equator and latitude 45


Homework Equations



circle: x^2 + y^2 = r^2



The Attempt at a Solution



so far, I have just been working on A. I took a cross section of the sphere from latitude 45 and up and drew it on a graph. I realized that if i revolved it around the y-axis that it would form the shape I need, a dome.

I found the radius using the circumference and set up my integral.
i have: pi * int(124500pi - y^2 dy after simplifying.

I think i'm all set to integrate and find the answer, but I cant figure out what to use for the upper and lower bounds. I thought i might try and use sin(45) or cos(45) or tan(45), in some way, but I cant really wrap my head around the problem from this point forward.
 

Answers and Replies

  • #2
tiny-tim
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… avoid using huge numbers … !

i have: pi * int(124500pi - y^2 dy after simplifying.

I think i'm all set to integrate and find the answer, but I cant figure out what to use for the upper and lower bounds. I thought i might try and use sin(45) or cos(45) or tan(45), in some way, but I cant really wrap my head around the problem from this point forward.

Hi xcgirl! :smile:

Hint: with big numbers like this, just put the radius = r throughout the calculation, and then put the number for r in at the end - you're much less likely to make a mistake (like forgetting to square something!) - and you won't have five-digit limits for the integral sign!

Yes, your approach seems fine.

Integrate over y, from y = r/√2 to r. :smile:

(If in doubt as to whether it's sin or tan, draw a diagram!)
 
  • #3
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when i do this, i am getting very large numbers, larger even then the actual volume of the earth. I'm not sure whats going on.
 
  • #4
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The integral you might be using, which seems to be

[tex]V = \pi\int_{\frac{r}{\sqrt{2}}}^r (124500\pi - y^2)dy[/tex]

doesn't look right. You may have simplified wrong; from where did you get [itex]124500\pi[/itex]?
 
  • #5
tiny-tim
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… let's see …

when i do this, i am getting very large numbers, larger even then the actual volume of the earth. I'm not sure whats going on.

Hi xcgirl! :smile:

Show us the integral you used, before putting any numbers in (ie just using r), so we can see what is going wrong. :smile:
 
  • #6
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http://www4a.wolframalpha.com/Calculate/MSP/MSP103319a0269e42g61e9e00000ga3250hih542d42?MSPStoreType=image/gif&s=35&w=114&h=48 [Broken]
This is the integral I used and I get 1.6376*10^12 miles cubed which doesn't seem right.
 
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  • #7
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If the earth were a cube 8000 miles on each side, its volume would be (8,000)3 mi3 = 512 x 109 mi3 = 5.12 x 1011 mi3. Being roughly spherical, the earth would fit inside such a box, so its volume would be less than this. That makes the value too big by maybe two orders of magnitude, since you're calculating the volume above 45 degrees N.
 

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