If the radius of a sphere is increased by $10\%$, by what percentage is its volume increased? 
Question: If the radius of a sphere is increased by $10\%$, by what percentage is its volume increased? Use Calculus.
Answer: It increases by approximately $33.1\%$.


I did the above question using calculus but my answer came out to be 30%. Here's the solution. What is wrong in it?

$V = \frac{4}{3} πr^3$
$dV = \frac 4 3  π \cdot 3  r^2 dr$
$dV  = 4 π r^2 dr$
since
$dr = \frac{10}{100} r$
therefore
$dV = 4  π  r^2 \cdot \frac{10}{100}  r$
$dV = \frac 4 {10} π r^3$
now
$dV/V = [ \frac 4{10}  πr^3] / [ \frac 4 3  πr^3]$
$dV/V = 3/10$
and
$100dV/V = \frac 3{10} \times 100 = 30\%  $

we have $$V_1=\frac{4}{3}\pi \cdot r^3$$ then we get $$V_2=\frac{4}{3}\pi\cdot r^3\left(\frac{11}{10}\right)^3$$ and we get $$\frac{V_2}{V_1}=\left(\frac{11}{10}\right)^3$$
