Maximum eror in calculated surface area

  • #1

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:
  • #2
Never mind, I figured it out.
  • #3
You are calculating the error in the measured volume - not in the measured surface area.

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