Question

6. The diameter of a sphere is measured to be 40cm. If an error of 0.02 cm is made in it
approximate errors in volume and surface area of the sphere.​

Answer 1

Answer:

Solved

Step-by-step explanation:

radius of the sphere = 20 cm

Δr= 0.01 cm

volume of the sphere $V= \frac{4}{3}\pi r^3$

we need to find error in calculating volume ΔV

$\Delta V=\frac{dV}{dr}\Delta r$

=$\frac{d\frac{4}{3}\pi r^3 }{dr}\Delta r$

$\Delta V= \frac{4}{3}\pi(3r^2)(0.01)$

=4/3π(20)^2×0.01

= 16.75 cm^2

Also surface area

S= 4πr^2

now we need to find ΔS

$\Delta S =\frac{dS}{dr}\Delta r$

$\Delta S =\frac{4\pi r^2}{dr}\Delta r$

ΔS = 8πrΔr

= 8π20×0.01 = 1.6π= 5.0265 cm^2