calculusconic sectionsdefinite integrals
I have this problem that I have worked out. Will someone check it for me? I feel like it is not correct. Thank you!
Rotate the graph of the ellipse about the $x$-axis to form an ellipsoid. Calculate the precise surface area of the ellipsoid.
$$\left(\frac{x}{3}\right)^{2}+\left(\frac{y}{2}\right)^{2}=1.$$
There is a nice equation describing the equation of an egg curve credit to Nobuo Yamamoto : $$ (x^2+y^2)^2 = ax^3 + \frac{3a}{10}xy^2, \tag{1} $$ where $0\leq x\leq a$, $a$ is the length of the major axis of symmetry for an egg.
In other words, we could get it by cutting a boiled egg in half and measure the distance from tip to the bottom. I just drew it in MATLAB, and the curve looks like the following for $a=1$:
I must say this curve fits pretty well with an egg. Now we have a curve, then the method of computing surface area by revolution, which is taught in Calculus II I believe, can be used to computing the surface. We just revolve the curve above around the egg's major axis of symmetry and get a surface, here is what looks like when we revolve it around the $x$-axis by degree $\pi$, we can get the lower half by revolve another $\pi$:
First we solve for $y$ in (1) when $y>0$: $$ y = \sqrt{\frac{3ax}{20} - x^2 + x\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} }, $$ The formula of computing surface area by revolution is: $$ S = 2\pi\int_0^a y\sqrt{1+\left(\frac{dy}{dx}\right)^2} \,dx,\tag{2} $$ The derivative is: \begin{align} \frac{dy}{dx} = \frac{1}{2\sqrt{\frac{3ax}{20} - x^2 + x\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} }} \left( \frac{3a}{20}- 2x+ \sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} + \frac{7ax}{20\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}}}\right), \end{align} Plugging $dy/dx$ and $y$ into (2), then you could use your favorite tool of numerical integration to perform the computing for you(Octave, MATLAB, Mathematica, etc).
A more tweakable/numerical/experimental approach:
As Ｊ. M. suggests in the comments, the shape looks like an egg, but is a real egg being approximated nicely by this curve? I guess the answer is that "it really depends on that specific egg"!
Let's say we still want to use surface of revolution to compute the surface area.
But this time, we handle it more numerically from the very start, instead of looking for a curve to fit one thing for all.
Two assumptions:
Now we want to compute the integral (2) using Simpson's rule or Trapezoidal rule, which is also taught in Calculus II in most colleges I believe. A remark is that $|f'|\to \infty$ when $x\to 0$ and $x\to a$, it would be much better if we use adaptive qudrature by putting more sample points near $0$ and $a$.
The steps are:
Boil an egg, cut it by half, hold it against a paper, use a pencil to outline its boundary (upper half is enough).
Draw the major axis, set it to be the $x$-axis, and measure its length $a$.
Choose $(n+1)$-sample points (including the end points) so that the points are equidistant to their neighbor on the curve. $n$ is chosen to be even, measure the distance to the major axis ($y$-coordinates) like the above figure.
Denote the sample point as $(x_i,y_i)$, $x_0=0$, $x_n = a$.
Approximate $$\frac{dy}{dx}\Big|_{x_i} \approx s_i = \frac{1}{2}\left( \frac{y_{i+1} - y_i}{x_{i+1} - x_i} + \frac{y_{i} - y_{i-1}}{x_{i} - x_{i-1}} \right).$$ For two end points: $$ s_0 = \frac{y_1 - y_0}{x_1 - x_0},\quad \text{ and }\quad s_n = \frac{y_n- y_{n-1}}{x_n - x_{n-1}}. $$ This step may be problematic, we can use other methods to approximate $dy/dx$: for example, approximating the curve by cubic spline using sample points $(x_i,y_i)$, but it would be beyond the content of college calculus.
Let $h = x_{i+1} - x_i = a/n$, approximate (2) by computing: $$ \frac{2\pi h}{3}\bigg[g(x_0)+2\sum_{j=1}^{n/2-1}g(x_{2j})+ 4\sum_{j=1}^{n/2}g(x_{2j-1})+g(x_n) \bigg], \quad \text{ where } g(x_i) = y_i \sqrt{1+s_i^2}. $$
Some results comparison:
Amzoti gave a link in his comments above that has two semi-empirical formulas: $$ S_1 = (3.155 − 0.0136a + 0.0115b)ab, \;\text{ and }\;S_2 = \left(0.9658\frac{b}{a}+2.1378\right)ab $$ where $a$ and $b$ are the length for major and minor axis of the real eggs. If there exists an egg's shape like (1), $a = 1$, and $b\approx 0.7242629$, the surface area computed by above formula is: $$ S_1 \approx 2.278215 \;\text{ and }\;S_2 \approx 2.054946. $$ Using the surface by revolution formula (2), we have: $$ S \approx 2.042087. $$
In this answer to a related question, it is shown that $$ Ax^2+Bxy+Cy^2+Dx+Ey+F=0 $$ is simply a rotated and translated version of $$ \small\left(A{+}C-\sqrt{(A{-}C)^2+B^2}\right)x^2+\left(A{+}C+\sqrt{(A{-}C)^2+B^2}\right)y^2+2\left(F-\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}\right)=0 $$ which says the semi-major axis is $$ \left[\frac{2\left(\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}-F\right)}{\left(A{+}C-\sqrt{(A{-}C)^2+B^2}\right)}\right]^{1/2} $$ and the semi-minor axis is $$ \left[\frac{2\left(\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}-F\right)}{\left(A{+}C+\sqrt{(A{-}C)^2+B^2}\right)}\right]^{1/2} $$
Best Answer
$\frac {x^2}{9} + \frac {y^2}{4} = 1\\ \frac {d}{dx}(\frac {x^2}{9} + \frac {y^2}{4} = 1)\\ \frac {2x}{9} + \frac {2y}{4}\frac {dy}{dx} = 0\\ \frac {dy}{dx} = - \frac {4 x}{9 y}\\ 2\pi \int_{-3}^{3} y\sqrt {1+(\frac {dy}{dx})^2}\ dx\\ 2\pi \int_{-3}^{3} \sqrt {y^2+ \frac {16}{81}x^2}\ dx$
Note you have the wrong limits of integration. Now you can substitute $y^2 = 4 - \frac 49 x^2$
$2\pi \int_{-3}^{3} \sqrt {4 - \frac {20}{81}x^2}\ dx\\ 2\pi \int_{-3}^{3} \frac {2}{9}\sqrt {81 - 5 x^2}\ dx$
Which gets you back where you were. Now you need to make a trig substitution $x = \frac 9{\sqrt 5} \sin \theta$ $ 2\pi \int_{-\arcsin \frac{\sqrt 5}3}^{\arcsin \frac{\sqrt 5}3} \frac {2}{9}\sqrt {81 - 81\sin^2\theta} (\frac 9{\sqrt 5} \cos \theta\ d\theta)\\ 2\pi \int_{-\arcsin \frac{\sqrt 5}3}^{\arcsin \frac{\sqrt 5}3} \frac {18}{\sqrt 5} \cos^2\theta\ d\theta\\ 2\pi \frac {9}{\sqrt 5}(\theta + \sin\theta\cos\theta)|_{-\arcsin \frac{\sqrt 5}3}^{\arcsin \frac{\sqrt 5}3}\\ \frac {36\pi}{\sqrt 5}\arcsin \frac{\sqrt 5}3 + 8\pi $