Maximum eror in calculated surface area

[Weave]

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.

 

[Weave]

Never mind, I figured it out.

 

[Dick]

You are calculating the error in the measured volume - not in the measured surface area.