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The question is as follows:

Write an expression for the volume of the spherical shell formed between two concentric spheres, the inner one of radius $r$, the outer one of radius $R$. Factor your answer so that it has the form $4 π · (trinomial) · (binomial)$. In this situation, what is the meaning of the binomial? What can be said about the value of the trinomial when the binomial has a very small value? Make a conjecture concerning the surface area of a sphere of radius $R$.

I know that the volume of the sphere with the inner radius of r would be $\frac{4}{3}\pi r^3$ and the volume of the sphere with the outer radius of R would be $\frac{4}{3}\pi R^3$. Therefore, the volume of the spherical shell formed between the concentric circles would be the difference between the outer sphere and the inner sphere--which, when mathematically written, would be $\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3$. That expression, after it's factored, would be $\frac{4}{3}\pi(R^3 - r^3)$. My question is how can I represent $(R^3 - r^3)$ the product of a trinomial and binomial. Any help will be greatly appreciated!

EDIT: I think that I may have found the answer to the question that I have presented above: $(R^2 + Rr + r^2)(R - r)$. Therefore, my complete answer to the volume of the spherical shell is $\frac{4}{3}(R^2 + Rr + r^2)(R - r)$.

I still don't know what the binomial is representing for this particular problem, nor its relation to the trinomial when the binomial value gets smaller. I think that I can use the help for understanding what the binomial and trinomial represents as a way to formulate a conjecture regarding the surface area of a sphere with radius R.

EDIT #2: The binomial, as I understand it, represents the thickness of the spherical shell formed by the concentric circles.

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    $\begingroup$ Since this is the kind of thing you have to just know, I don't mind telling you flat-out that one of the terms is $R-r$. You can find the other term by long division. $\endgroup$ – dfan Jan 27 '18 at 18:12
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    $\begingroup$ Factorize as $a^3-b^3=(a-b)(a^2+b^2+ab)$. $\endgroup$ – MalayTheDynamo Jan 27 '18 at 18:13
  • $\begingroup$ Thank you @MalayTheDynamo and @dfan! I have found the answer right after I posted the problem. However, would anyone of you help in guiding me in the questions that I posted in the third paragraph? $\endgroup$ – geo_freak Jan 27 '18 at 18:17
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Note that $$\frac{4}{3} \pi (R^2 + Rr + r^2)(R - r)$$

has three parts.

You have already figured out the $(R-r)$ part as the thickness of the spherical shell.

The trinomial part is very close to $ 3r^2$ in case that the thickness is very small.

Now when you multiply this $3r^2$ by $ \frac{4}{3} \pi$, you come up with $$ 4\pi r^2$$ which is the surface area of the shell( Approximate of course)

Thus The volume of the shell is approximately, the surface area of the shell multiplied by the thickness of the shell.

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  • $\begingroup$ How did you get that the trinomial part will be very close to $3r^2$? $\endgroup$ – geo_freak Jan 27 '18 at 18:35
  • $\begingroup$ Writing $r$ in place of $R$ as these are near to each othert $\endgroup$ – Narasimham Jan 27 '18 at 18:45
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    $\begingroup$ Later you can also find that surface area is $ 4 \pi r^2$ as differential coefft. of $ 4/3 \pi r^3$ with respect to $r$ $\endgroup$ – Narasimham Jan 27 '18 at 18:47
  • $\begingroup$ I see! Thank you so much for clarifying @Narasimham! $\endgroup$ – geo_freak Jan 27 '18 at 18:51
  • $\begingroup$ Thank you for your attention. $\endgroup$ – Mohammad Riazi-Kermani Jan 27 '18 at 18:53
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Let $V(r)$ be the volume of the sphere with radius $r$ and $A(r)$ its surface area.We have $$ V(r) = V(1)\cdot r^3$$ since concentric spheres are homothetic. In particular $$ V(R)-V(r) = V(1)(R-r)(R^2+Rr+r^2). $$ On the other hand $A(r) = A(1)\cdot r^2$ and $$ A(R) = \lim_{r\to R^-}\frac{V(R)-V(r)}{R-r} = \lim_{r\to R^-} V(1)(R^2+Rr+r^2)=3V(1)\,R^2.$$ This proves $A(1)=3\,V(1)$ and the same argument in dimension $n$ proves $A(1)=n\,V(1)$: to find the volume or the surface area of a Euclidean ball are equivalent problems. Remark: for $n=3$, the same can be shown by approximating a sphere with the union of a large number of cones and recalling that the volume of a cone is given by one third of the product between the height and the base area.

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  • $\begingroup$ Thank you for the answer that you have posted. I, unfortunately, am having a hard time understanding it since I am just taking Alg II/Trig for my maths class. Is it possible to explain it in more simpler terms? $\endgroup$ – geo_freak Jan 27 '18 at 18:28
  • $\begingroup$ @geo_freak: $A(R)$ is the derivative of $V(R)$ with respect to $R$ by elementary geometric considerations. In particular the ratio between $A(1)$ and $V(1)$ exactly equals the dimension of the space. $\endgroup$ – Jack D'Aurizio Jan 27 '18 at 18:33
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Write an expression for the volume of the spherical shell formed between two concentric spheres, the inner one of radius $r$, the outer one of radius $R$. Factor your answer so that it has the form $4 \pi \cdot (\text{trinomial}) \cdot (\text{binomial})$

$$ V(R, r) = \frac{4}{3}\pi (R^3-r^3) = \frac{4}{3}\pi(R-r)(R^2+r^2 + Rr) = 4 \pi \, t(R,r) \, b(R,r) $$ with \begin{align} t(R,r) &= \frac{1}{3}(R^2+r^2 + Rr) \\ b(R, r) &= R-r \end{align}

What can be said about the value of the trinomial when the binomial has a very small value?

In this case we have $r \approx R$ and we can estimate the volume by $$ V(R,r) \approx A(r) \, b(R,r) $$ where $A(r)$ is the area of a sphere with radius $r$. So we get $$ t(R,r) = \frac{1}{3}(R^2+r^2 + Rr) = \frac{V(R,r)}{4\pi b(R,r)} \approx \frac{A(r) \, b(R,r)}{4\pi b(R,r)} = \frac{A(r)}{4\pi} $$

Make a conjecture concerning the surface area of a sphere of radius R.

$$ A(t) \approx 4\pi\, t(R,r) = 4\pi \, \frac{1}{3}(R^2 + r^2 + Rr) \approx 4\pi \, \frac{1}{3}(r^2 + r^2 + r^2) = 4 \pi r^2 $$

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