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

$$SA=4\pi\(r^2$$ Eq1.
$$dV=8\pi\(rdr$$ Eq2.
$$c=2\pi\(r$$ Eq3.

The Attempt at a Solution

I solved for the radius by $$r=\frac{73}{2\pi}$$
I then plugged r into Eq2, I set $$dr= \frac{.5}{2\pi}$$
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.