## anonymous one year ago 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

1. anonymous

where did you get dy = 4πx^2*dx

2. mathmate

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.