Maximum eror in calculated surface area

  • Thread starter Weave
  • Start date
  • #1
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Homework Statement


The circumference of a sphere was measured to be 73.000 cm with a possible error of 0.50000 cm. Use linear approximation to estimate the maximum error in the calculated surface area


Homework Equations


[tex]SA=4\pi\(r^2[/tex] Eq1.
[tex]dV=8\pi\(rdr[/tex] Eq2.
[tex]c=2\pi\(r[/tex] Eq3.

The Attempt at a Solution


I solved for the radius by [tex]r=\frac{73}{2\pi}[/tex]
I then plugged r into Eq2, I set [tex]dr= \frac{.5}{2\pi}[/tex]
I get something like 134.98 and it is wrong.
 
Last edited:

Answers and Replies

  • #2
143
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Never mind, I figured it out.
 
  • #3
Dick
Science Advisor
Homework Helper
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You are calculating the error in the measured volume - not in the measured surface area.
 

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