1
$\begingroup$

I'm a high school senior doing the AP Calc course, and recently I studied surface area of revolutions. As background research for a class project, I tried to look for things found in nature that roughly involve revolutions, and shortlisted eggs and lemons. But eggs were already analysed, and lemons seemed more challenging.

So now I'm trying to find an approximation for the surface area of a lemon.

From research, I've found that a lemon takes the approximate shape of a prolate spheroid. This is what that shape looks like:

Prolate Spheroid

I know that the equation for the surface area of revolution is:

$S = 2\pi\int_a^b f(x)\sqrt{1+\left(\frac{dy}{dx}\right)^2} \,dx$

I can try finding the side profile of a lemon, model it into a function, and then apply the equation above from 0 to pi. However, this doesn't look like the best way to me, because I'm not sure how to find an equation to describe a lemon's shape. Plus, the lemons in different countries, such as India, are of a different shape.

I'm looking for a different way to model the equation describing a lemon's cross-sectional shape. My knowledge about elliptic integrals is limited, but I'm willing to do research to find a way which involves mathematics beyond my high-school level.

[Update] After research, I realized rather than going for an ellipse rotated around the axis, I could go for a lens shape too. A bulbous lens might approximate the ends of the lemon too.

$\endgroup$
  • $\begingroup$ I do not know the shape of Indian lemons but the surface area of an ellipsoid is given by an elliptic integral and there are may efficient algorithms for the numerical evaluation of such objects, through the AGM mean or variations of it. $\endgroup$ – Jack D'Aurizio Aug 21 '17 at 12:56
  • 1
    $\begingroup$ @JackD'Aurizio No need to go that far. Prolate spheroid is an ellipsoid of revolution, and its surface area is expressed in elementary functions. Then again, if the lemons are of the wrong shape, we might need an entirely different approximation. $\endgroup$ – Ivan Neretin Aug 21 '17 at 13:02
  • $\begingroup$ @IvanNeretin: do you consider elliptic integrals as elementary functions? en.wikipedia.org/wiki/Ellipsoid#Surface_area $\endgroup$ – Jack D'Aurizio Aug 21 '17 at 13:05
  • 2
    $\begingroup$ @JackD'Aurizio Nope, but our ellipsoid is not a tri-axial one. When two axes are equal, the elliptic integrals just seem to go away somehow. en.wikipedia.org/wiki/Spheroid#Area $\endgroup$ – Ivan Neretin Aug 21 '17 at 13:11
  • $\begingroup$ @IvanNeretin: oh, nice, I was not aware of such simplification, thanks. $\endgroup$ – Jack D'Aurizio Aug 21 '17 at 13:12
0
$\begingroup$

The equations for the surface area of spheroids have been pointed out in the comments. Not knowing the cross-section of the Indian lemon I'll point out a more general approach to the surface area. Assuming that that the lemon is a body of revolution, we can find the surface are from Pappus's $(1^{st})$ Centroid Theorem: the surface area $A$ of a surface of revolution generated by rotating a plane curve $C$ about an axis external to $C$and on the same plane is equal to the product of the arc length $s$ of $C$ and the distance d traveled by its geometric centroid. Simply put, $S=2\pi RL$, where $R$ is the normal distance of the centroid to the axis of revolution and $L$ is curve length. The centroid of a curve is given by

$$\mathbf{R}=\frac{\int \mathbf{r}ds}{\int ds}=\frac{1}{L} \int \mathbf{r}ds$$

In the complex plane, the surface area of a curve rotated about the $x$-axis is given by

$$S=2\pi\int \Im\{z\}|\dot z| du,\quad z=z(u)$$

As an example, we can find the surface area of an arbitrary ellipse of revolution given by

$$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$$

Thus,

$$ \begin{align} &z=a\cos\theta+ib\sin\theta,\quad \theta\in[0,\pi]\\ &\dot z=-a\sin\theta+ib\cos\theta\\ &|\dot z|=\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}=b\sqrt{\left(\frac{a^2-b^2}{b^2}\right)\sin^2\theta+1}\\ \end{align} $$

So that finally

$$ \begin{align} S &=2\pi\int_0^{\pi}b^2\sin\theta\sqrt{\left(\frac{a^2-b^2}{b^2}\right)\sin^2\theta+1}\\ &=2\pi b^2\left[1+\frac{(1+A)\tan^{-1}\sqrt{A}}{\sqrt{A}}\right],\quad A=\left(\frac{a^2-b^2}{b^2}\right) \end{align} $$

This has been tested numerically for arbitrary $a$ and $b$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.