Area Enclosed By A Curve

Geometry Level 5

Let $A$ denote the area of the region enclosed by the curve $5x^2 + 6xy + 2y^2 +7x + 6y +6 = 0$ . Find the value of $\dfrac{10A}\pi$ .

This is my first problem! I would like to thank @Andrew Ellinor for the encouragement.

The answer is 5.

2 solutions

Otto Bretscher
May 20, 2016

This solution is based on joint work with Comrade @Pi Han Goh

Completing the squares, we find $2\left(y+\frac{3x}{2}+\frac{3}{2}\right)^2+\frac{1}{2}(x-2)^2-\frac{1}{2}=0$ . The transformation $u=x-2, v=\frac{3x}{2}+y+\frac{3}{2}$ preserves area, being a shear followed by a translation. The ellipse $2u^2+\frac{v^2}{2}=\frac{1}{2}$ or $4u^2+v^2=1$ has semi-axes $a=\frac{1}{2}$ and $b=1$ and its area is $ab\pi=\frac{\pi}{2}$ . The answer is $\boxed{5}$

Never knew this simple method. Thanks a lot.

Niranjan Khanderia - 4 years, 3 months ago

FWIW, I consider this the "standard" way to find the area of an ellipse, and then the determinant approach merely masks all of the calculations / understanding that is happening here.

Calvin Lin Staff - 5 years ago

Thanks.... agreed!

Otto Bretscher - 5 years ago

being a shear followed by a translation

What does this mean? Did you do Jacobian as well?

Pi Han Goh - 5 years ago

I'm trying to make things as simple as possible, so, no, I'm not using the Jacobian.

Shear mappings are described here. The one we are using is a vertical shear. Shears preserve area by Cavalieri's principle.

Otto Bretscher - 5 years ago

Oh Em Goodness! The more I know, the more I don't know!!!

Pi Han Goh - 5 years ago

@Pi Han Goh – We are all in the same boat, Comrade!

Otto Bretscher - 5 years ago

@Pi Han Goh – Comrade @Pi Han Goh , what do you think about my comments to this problem ? I'm afraid the problem and the solution make no sense at all.

Otto Bretscher - 5 years ago

@Otto Bretscher – It is not meant to be a Riemann integral, but it is a Riemann-Stieltjes integral .

Pi Han Goh - 5 years ago

@Pi Han Goh – The poser of the problem should state that then. I will just ask what notion of an integral he is using. I thought it was some fancy Dirac Delta thing, but that one is tricky when the discontinuities are at the end points.

Otto Bretscher - 5 years ago

@Otto Bretscher – Good point. Let me file a report.

Pi Han Goh - 5 years ago

@Pi Han Goh – I'm asking: What kind of an integral are you using? Maybe he figures it out himself. And then we can ask him to show that integration by parts works in that case ;)

Otto Bretscher - 5 years ago

@Otto Bretscher – $\int_0^2 (x^2+1) \, d\lfloor x \rfloor = (x^2+1) \lfloor x \rfloor |_0^2 - \int_0^2 \lfloor x \rfloor \, d(x^2+1) = 10 - \int_1^2 \, d (x^2 + 1) = 10 - \int_1^2 2x \, dx = 10 - [ x^2 ]_1^2 = 10 - (2^2-1^2)=\boxed7$

Or because the integral, $\displaystyle \int_a^b f(x) \, d(g(x)) := \sum_{i=1}^n f(z_i) (g(x_i) - g(x_{i-1})$ , with $g(x) = \lfloor x \rfloor$ . For the last interval, $g$ jumps from 1 to 2 (step function). So the integral is $f(1) + f(2) =7$ .

Pi Han Goh - 5 years ago

@Pi Han Goh – If he does indeed use the Riemann-Stieltjes integral, then the answer is trivial, by definition: It's just $f(1)+f(2)$ , where $f(x)=x^2+1$ ... weird problem

Otto Bretscher - 5 years ago

Pi Han Goh
May 19, 2016

We need to convert the equation of the curve into the assciated quadratic form, $A X^2 + BXY + CY^2 + D = 0$ , where $X$ and $Y$ are the new axes. In other words, we want to eliminate the linear terms from the equation of the curve, $5x^2 + 6xy + 2y^2 + 7x+6y+6 = 0$ .

Let $x = X - a$ and $y = Y-b$ , then we have

$\begin{aligned} && 5x^2 + 6xy + 2y^2 + 7x + 7y + 6 \\ &=& 5(X-a)^2 + 6(X-a)(Y-b) + 2(Y-b)^2 + 7(X-a)(Y-b) + 6 \\ &=& 5X^2 + 6XY + 2Y2 + X(-10a-6b+7) + Y(-6a-4b+6) + (5a^2 + 6ab - 7a-6b + 6 + 2b^2) \end{aligned}$

Setting the coefficients of the terms of $X$ and $Y$ to 0, we can find the values of $a$ and $b$ by solving the system of simultaneous equations:

$\begin{cases} -10a - 6b + 7=0 \\ -6a - 4b + 6 = 0 \end{cases}.$

Which we will get $a = -2, b = \dfrac92$ . Substituting them into the equation of the curve in question, we have

$5x^2 + 6xy + 2y^2 + 7x + 7y + 6 = 5X^2 + 6XY + 2Y^2 - \dfrac12.$

So $A$ can be calculated as the area of the region enclosed by the curve $5X^2 + 6XY + 2Y^2 - \dfrac12 = 0$ , or equivalently (removing the fraction): $10X^2 + 12XY + 4Y^2 - 1 = 0$ .

Because the eigen decomposition of associated quadratic form $B:= \begin{bmatrix}{10} && {6} \\ {6} && {4}\end{bmatrix}$ can be written as $B = U \Lambda U^T$ , where $U$ is an orthonormal matrix and $\Lambda$ is a diagonal matrix consisting of eigenvalues of $A$ , and noting that the rotation does not change the area.

Since $\Lambda$ is diagonal, it is equivalent to $\lambda_1 \zeta_1^2 + \lambda_2 \zeta_2^2 = 1$ , which is an ellipse in a standard form and its area $A$ is

$A = \dfrac{\pi}{\sqrt{\lambda_1 \lambda_2}} = \dfrac{\pi}{\sqrt{\det (B)}} \;$

where $\det (B) = \begin{vmatrix}{10} && {6} \\ {6} && {4}\end{vmatrix} = 10\times4- 6^2 = 4$ . Hence, $A = \dfrac{\pi}{\sqrt4} = \dfrac{\pi}2$ . And so, the desired answer is $\dfrac{10A}{\pi} =\boxed5$ .

Moderator note:

A lot of work that you're doing here is obscured by the notation used. Essentially, we want to determine the length of the major and minor axis of the ellipse, and then the area is $\pi a b$ .

@Otto Bretscher , any improvements for my solution?

Pi Han Goh - 5 years ago

Oh yes, this is very nicely done and well explained, Comrade! Wunderbar! (+1)

There are some optional short cuts here and there, when you are in a hurry. In the equation $AX^2+BXY+CY^2+D=0$ , the constant $D$ is simply the minimal value of the LHS. As we have seen in other problems (I'm sure you remember those!), this minimum can sometimes be found most quickly by completing the squares. In our case:

$2\left(y+\frac{3x}{2}+\frac{3}{2}\right)^2+\frac{1}{2}(x-2)^2-\frac{1}{2}=0$

so $D=-\frac{1}{2}$ . Your method takes a bit longer, but it explains better what is going on.

Small typo, just to prove that I read it: There is a square root missing in the penultimate line, $\sqrt{\det(B)}$

Otto Bretscher - 5 years ago

If you really want to be lazy, you could just write it as $5X^2 + 6XY+ 2Y^2 - \min(5X^2+6X^2+2Y^2) = 0$ since we only care about quadratic form and nothing else.

Thanks. I've caught and fixed all my typos already.

Ah yes, completing the squares! Underrated technique. With $2\left(y+\frac{3x}{2}+\frac{3}{2}\right)^2+\frac{1}{2}(x-2)^2-\frac{1}{2}=0$ , we can let $u = y + \dfrac32x + \dfrac32, v = x-2$ . By Jacobian, the determinant $J = \begin{vmatrix}{\frac{\partial u}{\partial x}} && {\frac{\partial u}{\partial y}} \\ {\frac{\partial v}{\partial x}} && {\frac{\partial v}{\partial y}}\end{vmatrix} = 1$ . So the area is preserved. And $A$ is equal to the area of the ellipse $2u^2 + \dfrac{v^2}2 = \dfrac12 \Leftrightarrow \dfrac{u^2}{(1/2)^2} + v^2 = 1$ . Hence, $A = \pi \times \dfrac12 \times1 = \dfrac\pi2$ .

Do you have an alternative/better solution? Like "area by integration" or "parametric equations"?

Pi Han Goh - 5 years ago

@Pi Han Goh – Don't get too lazy now ;) We need to know the value of the minimum to find the area (and it's not just the minimum of $5X^2+6XY+2Y^2$ , which is zero, but of the whole thing).

Your remarks on completing the squares are right on target! The transformation is a shear-translation, preserving area.

Otto Bretscher - 5 years ago

@Otto Bretscher – Oh right. I was being lazy wrongly. Thanks for everything, comrade!

Pi Han Goh - 5 years ago

@Pi Han Goh – Shall I write an alternate solution ("based on joint work with Comrade Pi Han Goh"), using the "underrated technique" of completing the squares? It's basically in our comments now, but it may be worth organizing it into a three-line solution.

Otto Bretscher - 5 years ago

@Otto Bretscher – Yes plzzzzzz!

Pi Han Goh - 5 years ago

@Pi Han Goh – This solution is based on joint work with Comrade @Pi Han Goh

Otto Bretscher - 5 years ago

@Otto Bretscher – No! Post it as a solution, not a comment.

Pi Han Goh - 5 years ago

@Pi Han Goh – Well, your link yesterday took me directly to the solution (for some reason I had not previously seen the problem), so, I'm not entitled to write a solution ;) Maybe @Calvin Lin or another moderator will be kind enough to convert my comment into a solution.

Otto Bretscher - 5 years ago

Area Enclosed By A Curve

This is my first problem! I would like to thank @Andrew Ellinor for the encouragement.

The answer is 5.

2 solutions

Moderator note:

0 pending reports