If volume of a sphere increases by 72.8%, what is change of its surface area?

Question

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?

If you take the ratio of the "new" volume vs "old", you will find $R=r1.728^{1/3}$ where R is the new radius, r the old radius. — David P, Oct 14 '19 at 14:58
@copper.hat it was a question I found in a book that dealt with derivatives — Techie5879, Oct 14 '19 at 15:35
Looking at Quanto's answer, you have $S = C , V^{2/3}$ so that $dS = 2/3 , C , V^{-1/3} , d V$ and $dS / S = 2/3 , dV / V$. — jim, Oct 14 '19 at 23:16

score 5 · Accepted Answer · answered Oct 14 '19 at 14:54

5

Recall that

volume $V= \frac43 \pi r^3$
surface area $S= 4 \pi r^2$

then assume

$V+\Delta V=\frac43 \pi (r+\Delta r)^3 =1.728 V=1.728\cdot \frac43 \pi r^3$

and find $\Delta r$.

answered Oct 14 '19 at 14:54

user

154,566

Quanto · Answer 2 · 2019-10-16T16:26:59.867

5

Use $V= \frac43 \pi r^3$ and $S= 4 \pi r^2$ to establish

$$S=(36\pi)^{1/3}V^{2/3}$$

With the new volume $V'=1.728V=1.2^3V$, the new surface is

$$S'=(36\pi)^{1/3}(1.2^3V)^{2/3}=(1.2)^2S=1.44S$$ Thus, the surface increases 44%, exactly.

edited Oct 16 '19 at 16:26

answered Oct 14 '19 at 15:13

Quanto

97,352

Cute observation. I guess $M= 1.728 = \frac {1728}{1000}=\frac{12^3}{10^3}=\frac {12}{10}^3$ and so $M^{\frac 23} = \frac {12}{10}^2$. And FWIW the radius has increased from $1$ to $1.2$ or $20$. – fleablood Oct 14 '19 at 15:28
@fleablood - Yeah, the question is designed this way on purpose, anticipating that some would miss – Quanto Oct 14 '19 at 15:31

score 1 · Answer 3 · answered Oct 14 '19 at 14:57

Hint: A sphere of radius $r$ has volume $V=\frac43\pi r^3$ and surface area $A=4\pi r^2$. So $r=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt{\frac A{4\pi}}$. If the volume changes from $V$ to $1.728V$, then the radius changes from $r$ to ... and in turn the area changes from $A$ to ...

fleablood · Answer 4 · 2019-10-14T15:21:09.527

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 $A=\int_{-r}^4 2\sqrt{r^2 - x^2}dx = \pi r^2$

score 0 · Answer 5 · answered Oct 14 '19 at 15:01

0

As @user170231 siad, $r$ increases by $1.728^{1/3}$ and then $A$ increases by $1.728^{2/3}$, So the final answer will be $100(1.728^{2/3}-1) \%$.

answered Oct 14 '19 at 15:01

Ali Ashja'

2,786
8
12

If volume of a sphere increases by 72.8%, what is change of its surface area?

5 Answers5