Question

The radius of a sphere is claimed to be 3.0 inches, correct to within 0.01 inch. Use linear approximation to estimate the resulting error, measured in cubic inches, in the volume of the sphere.

± 0.000001
± 0.36π
± 0.036π
± 0.06

Answer 1

where did you get dy = 4πx^2*dx

Answer 2

Volume of a sphere of radius r is \(V=\frac{4}{3}\pi r^3 \)
\(dV/dr = \frac{4}{3}\pi (3)r2 = 4\pi r^2\), or
\(dV = \frac{4}{3}\pi (3)r2~dr = 4\pi r^2~dr\)
where dr=0.01" and r=3".
If you think of the situation physically,
4 pi r^2 is the surface area of the sphere. Multiply that by the error thickness will give you the error in volume.