If volume of a sphere increases by 72.8%, what is change of its surface area? If the volume of a sphere is increased by 72.8% what would be the change in surface area?
I'm trying to solve the problem using application of derivatives. I noticed that on differentiating the formula for volume of a sphere we directly end up with that of the surface area. How to approach this link?

Set up a formula that sets up Volume in terms of Area.
As Volume is determined by $V= \frac 43\pi r^3$  and Surface Area of a Sphere is $SA= 4 \pi r^2$ then $SA = 4\pi r^2 = 4\pi{\sqrt[3]{\frac {3V}{4\pi}}}^2$
So if volume is increased by $1.728$ then surface area is increased from $4\pi{\sqrt[3]{\frac {3r}{4\pi}}}^2$ to   $4\pi{\sqrt[3]{\frac {3*1.728r}{4\pi}}}^2$
And the proportional increase is $\frac{4\pi{\sqrt[3]{\frac {3*1.728r}{4\pi}}}^2}{4\pi{\sqrt[3]{\frac {3*1.728r}{4\pi}}}^2}=1.782^{\frac 23}$ 
And percentage increase is $100(1.782^{\frac 23}-1)$
The real question, I suppose, is how to get the formulas for volumes and surface areas in the first place.
If we look at cross section circles of  a sphere at points $x: -r\le x \le 4$ along the diameter of the sphere, these cross section circles have radii of $R = \sqrt{r^2 - x^2}$.
An area of one of these circles is $\pi R^2=\pi(r^2 - x^2)$   and the circumference of a circle is $2\pi R= 2\pi\sqrt{r^2 - x^2}$ 
So the volume of a sphere is $V= \int_{-r}^r \pi(r^2 - x^2)dx=\frac 43\pi r^3$ and the surface area of a sphere is $\int_{-r}^r2\pi\sqrt{r^2 - x^2}dx= 4\pi r^2$
.....
Well we are at it the area of circle is determine  by $A=\int_{-r}^4 2\sqrt{r^2 - x^2}dx = \pi r^2$
