]> Exercise 2

# Applications, Exercise 2

## Question

The two ellipses with a>b>0 are

$\begin{array}{l}\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\text{ }\left(\text{horizontal}\right)\\ \frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1\text{ }\left(\text{vertical}\right)\end{array}\right\}\text{and}$

We are interested in the central enclosed area.

1. Make a sketch and mark this area.
2. Find the intersection points.
3. Calculate the area.

## Reminder

.... one of the functions, say y1(x), should be greater than the second y2(x) for any point of the interval; otherwise we'll have intersection points inside the interval. From the definition of area under a curve, it follows that the enclosed area S is the difference of the integrals of the functions. Formally this can be written as

 $\begin{array}{l}\text{for}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\text{interval}\text{ }\left[a,b\right]\\ \left\{\begin{array}{l}{y}_{1}\left(a\right)={y}_{2}\left(a\right)\\ {y}_{1}\left(b\right)={y}_{2}\left(b\right)\end{array}\\ \text{for}\text{\hspace{0.17em}}x\ne a\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}x\ne b:\\ {y}_{1}\left(x\right)>{y}_{2}\left(x\right)\ge 0\\ S=\underset{a}{\overset{b}{\int }}{y}_{1}\text{d}x-\underset{a}{\overset{b}{\int }}{y}_{2}\text{d}x=\\ =\underset{a}{\overset{b}{\int }}{y}_{1}\text{d}x+\underset{b}{\overset{a}{\int }}{y}_{2}\text{d}x\end{array}\right\}$ (2.2.2.1)

The last row of (2.2.2.1) can be interpreted as the integral, whose path is clockwise along the loop, defined by the appropriate functions.

... what happens if there are points of the loop with negative  y  values? .... it does not matter: the result of the integration gives the enclosed area with positive sign, as far as the clockwise sense is observed.

## Introductory remark

The calculation of the area can be done in different ways. As it is explained here the area is divided in 8 equal segments consisting of a triangle and an area integrated on an elliptic arc, done directly on the canonical form. It could be done by use of the parametric form. The area could be calculated also without the need of a triangle, but just by integration on the elliptic arcs.

The way the calculation is done is of no importance as far as it is correct.

## Part 1

Solution of question 1

1. The Fig. sketch shows the area that should be calculated.

## Parts 2-3

Solution of question 2

1. The two ellipses can be rewritten as $\begin{array}{l}{a}^{2}{x}^{2}+{b}^{2}{y}^{2}={a}^{2}{b}^{2}\\ {b}^{2}{x}^{2}+{a}^{2}{y}^{2}={a}^{2}{b}^{2}\end{array}\right\}$ We can find the x coordinate of the intersection point by multiplying the upper equation by a², the lower by b² and subtract. The result is

$\begin{array}{l}\left({a}^{4}-{b}^{4}\right){x}^{2}={a}^{2}{b}^{2}\left({a}^{2}-{b}^{2}\right)\\ {x}^{2}={a}^{2}{b}^{2}\frac{\left({a}^{2}-{b}^{2}\right)}{\left({a}^{4}-{b}^{4}\right)}\end{array}$
2. Since $\left({a}^{4}-{b}^{4}\right)=\left({a}^{2}-{b}^{2}\right)\left({a}^{2}+{b}^{2}\right)$ and since $a\ne b$ we obtain from part 2 after taking square root $x=±\frac{ab}{\sqrt{{a}^{2}+{b}^{2}}}$ Because of the xy symmetry we obtain also $y=±\frac{ab}{\sqrt{{a}^{2}+{b}^{2}}}$ The signs of x and y are not correlated, therefore we have 4 intersection points, as it can be seen also at the Fig. sketch.

## Parts 4-8

Solution of question 3

1. The calculation of the area is done here by dividing the total area in 8 equal parts. The area of the part is obtained by adding the area of a triangle to the area obtained by integrating under an arc of the vertical ellipse. This is explained in the Fig. sketch

2. The triangle (from part 4) is right-angled and his area is $S\left(\text{triangle}\right)=\frac{|xy|}{2}=\frac{{a}^{2}{b}^{2}}{2\left({a}^{2}+{b}^{2}\right)}$ where x and y denote the coordinates of an intersection point (see part 3).

3. If we denote by xi the abscissa of the intersection point (part 3) ${x}_{i}=\frac{ab}{\sqrt{{a}^{2}+{b}^{2}}}$
then the area under an arc of the vertical allipse is

$S\left(\text{ellipse}\right)=\underset{{x}_{i}}{\overset{b}{\int }}y\text{d}x=a\underset{{x}_{i}}{\overset{b}{\int }}\sqrt{1-\frac{{x}^{2}}{{b}^{2}}}\text{d}x$

4. which becomes $\begin{array}{l}S\left(\text{ellipse}\right)=\frac{a}{b}\int \sqrt{{b}^{2}-{x}^{2}}\text{d}x=\\ =\frac{a}{b}\left[\frac{x\sqrt{{b}^{2}-{x}^{2}}}{2}+\frac{{b}^{2}}{2}\text{asin}\left(\frac{x}{b}\right)\right]\underset{{x}_{i}}{\overset{b}{|}}\end{array}$ by use of the table of integrals and after the substitution of xi from part 6 we obtain

$\begin{array}{l}S\left(\text{ellipse}\right)=\\ =\frac{a}{2b}\left[{b}^{2}\text{asin}\left(1\right)-\frac{ab}{\sqrt{{a}^{2}+{b}^{2}}}\sqrt{{b}^{2}-\frac{{a}^{2}{b}^{2}}{{a}^{2}+{b}^{2}}}-{b}^{2}\text{asin}\left(\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}\right)\right]=\\ =\frac{ab}{2}\left[\frac{\pi }{2}-\text{asin}\left(\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}\right)-\frac{ab}{{a}^{2}+{b}^{2}}\right]\end{array}$

5. We have finally the required area

$\begin{array}{l}S=8\left[S\left(\text{triangle}\right)+S\left(\text{ellipse}\right)\right]=\\ =4ab\left[\frac{\pi }{2}-\text{asin}\left(\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}\right)\right]\end{array}$

## Final remark

It is advisable to check the result, in particular when the calculation is long and a direct verification is not convenient. In this case e.g. one can make the following tests:

• If $a\to b=r$, the result should be $S\to \pi {r}^{2}$, as can be easily seen.
• If $a\to \infty$, the result should be $S\to 4{b}^{2}$. This is a bit more complicated since an expression of the type $0\infty$ is obtained and one has to apply the L'Hôpital rule once.

## Score

By parts:

Parts 1 and 7 are worth 2 point each.
Parts 2,3,4,5,6,8 are worth 1 point each

By questions:

Questions 1 and 2 are worth 2 points each.
Question 3 is worth 6 points.