anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
where did you get dy = 4πx^2*dx
mathmate
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.