Find the approximate error in the volume of the sphere if the radius of the sphere is measured to be 6cm with an error of 0.07cm

The correct option is 10.08π cm³.

To find the approximate error in the volume of a sphere, we can use differentials. Let us denote the radius of the sphere as $r$ and its volume as $V$.

The volume $V$ of a sphere of radius $r$ is given by the formula:
$V=\frac{4}{3}\pi r^3$

The approximate error in the volume, denoted by $dV$, can be calculated using the differential of $V$ with respect to $r$:
$dV=\frac{dV}{dr}\cdot dr$

Differentiating $V$ with respect to $r$, we get:
$\frac{dV}{dr}=\frac{4}{3}\pi \cdot 3r^2=4\pi r^2$

Therefore, the expression for the approximate error in volume is:
$dV=4\pi r^{2}dr$

From the given data:
- The measured radius of the sphere, $r=6\text{ cm}$
- The error in measuring the radius, $dr=0.07\text{ cm}$

Substituting these values into the differential equation:
$dV=4\pi \cdot {\left(6\right)}^2\cdot 0.07$

$dV=4\pi \cdot 36\cdot 0.07$

$dV=144\pi \cdot 0.07$

$dV=10.08\pi {\text{ cm}}^3$

Thus, the approximate error in the volume of the sphere is 10.08π cm³.