Browse Questions

# Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3cm and 3.0005cm,respectively.

Toolbox:
• If $y = f(x)$, then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
• $\bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
Volume of an hollow spherical shell = $\large\frac{4}{3}\pi (r_2^3-r_1^3)$
External radii $r_2=3.005 cm$
Internal radii $r_1=3cm$
$v=\large\frac{4}{3}\pi (r_2^3-r_1^3)$
Let us find the value of $(3.005)^3$
Let y = $f(r)=(3.005)^3$
Let $x=3\: and \: x+\Delta x=3.005$
$\therefore \Delta x = 3.005-3=0.005$
Hence $x = 3\: and \: \Delta x=0.005$
for $x=3$ we have
$y=(3)^3$
$=27$
$dx=\Delta x = 0.005$
Step 2
$y=x^3$
Differentiating on both sides. w.r.t $x$,
$\large\frac{dy}{dx}=3x^2$
$= dy=3x^2.dx$
Substituting for $x$ and $dx$ we get
$dy= \Delta y=3(3)^2 \times 0.005$
$=27 \times 0.005$
$\therefore dy=\Delta y = 0.135$
Hence $(3.005)^3=y+ \Delta y$
$= 27+.135$
$27.135$
Hence the volume of the hollow spherical shell is
$v = \large\frac{4}{3} \times \pi \times (27.135-27)$
$= \large\frac{4}{3} \times \pi \times 0.135$
$= 0.018 \pi cm^2$