Chapter 8.2, Problem 56E

a.

To determine

To find the equations that define the upper and lower semi ellipses as function of x .

Answer to Problem 56E

The equations that define the upper and lower semi ellipses as function of x is

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

Explanation of Solution

Given information:

The given equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , the length of major axis is 2a and the length of minor axis is 2b .

Calculation:

Solve the equation for y .

  $\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{y^2}{b^2}=1-\frac{x^2}{a^2}\left[\text{add}\left(-\frac{x^2}{a^2}\right)\text{ each side}\right]\\ y^2=b^2\left(1-\frac{x^2}{a^2}\right)\left[\text{multiply each side by}\left(b^2\right)\right]\\ y=\pm b\sqrt{\left(1-\frac{x^2}{a^2}\right)}\end{array}$

Therefore,

The equations that define the upper and lower semi ellipses as function of x is

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

b.

To determine

To write an integral expression that gives the area of the ellipse.

Answer to Problem 56E

The integral expression for the area of ellipse is $2{\displaystyle \underset{-a}{\overset{a}{\int }}b\sqrt{\left(1-\frac{x^2}{a^2}\right)}}$

Explanation of Solution

Given information:

The given equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , the length of major axis is 2a and the length of minor axis is 2b .

Calculation:

Since the ellipse is symmetric abut x axis and centred at the origin. Therefore area of half ellipse can be calculated as integrating y as a function of x in the range of major axis that is $\left(-a,a\right)$ . Hence the integral for area of the half ellipse,

  ${\displaystyle \underset{-a}{\overset{a}{\int }}b\sqrt{\left(1-\frac{x^2}{a^2}\right)}}$

And for the area of ellipse multiply the area of half ellipse by 2, as the ellipse is symmetric.

  $2{\displaystyle \underset{-a}{\overset{a}{\int }}b\sqrt{\left(1-\frac{x^2}{a^2}\right)}}$

Therefore,

The integral expression for the area of ellipse is $2{\displaystyle \underset{-a}{\overset{a}{\int }}b\sqrt{\left(1-\frac{x^2}{a^2}\right)}}$

c.

To determine

To find the area of ellipses for various lengths of a and b .

Answer to Problem 56E

For $a=3,b=1$ the area of the ellipse centred at the origin is 9.425

For $a=4,b=2$ the area of the ellipse centred at the origin is 25.132

Similarly for different values of a and b area of the ellipses can be evaluated.

Explanation of Solution

Given information:

The given equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , the length of major axis is 2a and the length of minor axis is 2b .

Calculation:

Consider $a=3,b=1$

Then the area of the ellipse centred at the origin.

  $\begin{array}{l}A=2{\displaystyle \underset{-3}{\overset{3}{\int }}\sqrt{\left(1-\frac{x^2}{3^2}\right)}}\\ A=9.425\text{ use NINT}\end{array}$

Again consider $a=4,b=2$

Then the area of the ellipse centred at the origin.

  $\begin{array}{l}A=2{\displaystyle \underset{-4}{\overset{4}{\int }}2\sqrt{\left(1-\frac{x^2}{4^2}\right)}}\\ A=25.132\text{ use NINT}\end{array}$

Similarly for different values of a and b area of the ellipses can be evaluated.

d.

To determine

To write the form of areas found in above parts.

Answer to Problem 56E

The form of area is $A=ab\pi$

Explanation of Solution

Given information:

The given equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , the length of major axis is 2a and the length of minor axis is 2b .

Calculation:

First write the integral obtained for calculation of area of ellipse.

  $2{\displaystyle \underset{-a}{\overset{a}{\int }}b\sqrt{\left(1-\frac{x^2}{a^2}\right)}}$

Now integrate above expression.

  $\begin{array}{l}A=2{\displaystyle \underset{-a}{\overset{a}{\int }}b\sqrt{\left(1-\frac{x^2}{a^2}\right)}}\\ A=2b{\displaystyle \underset{-a}{\overset{a}{\int }}\sqrt{\left(1-\frac{x^2}{a^2}\right)}}\\ \text{substitute }x=a\sin \left(u\right)\\ A=2b{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}a{\cos }^2\left(u\right)}du\\ A=2ab{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\cos }^2\left(u\right)}du\\ A=2ab{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{1+\cos \left(2u\right)}{2}}du\\ A=ab\left({\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}1}du+\mathrm{c}{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\cos \left(2u\right)}du\right)\\ A=ab\pi \end{array}$

Therefore,

The form of area is $A=ab\pi$

e.

To determine

To proof that the simple area formulae of ellipse is same as the integral formula of area of the ellipse.

Answer to Problem 56E

The graph of $y=f(2x)-1$ is obtained by shrinking the graph of $f$ by a factor of $\frac{1}{2}$ then shift downward by 1 unit.

Explanation of Solution

Given information:

The function is $y=f(2x)-1$

Proof:

The simple formula of area of ellipse centred at origin is $A=ab\pi$

First write the integral obtained for calculation of area of ellipse.

  $2{\displaystyle \underset{-a}{\overset{a}{\int }}b\sqrt{\left(1-\frac{x^2}{a^2}\right)}}$

Now integrate above expression.

  $\begin{array}{l}A=2{\displaystyle \underset{-a}{\overset{a}{\int }}b\sqrt{\left(1-\frac{x^2}{a^2}\right)}}\\ A=2b{\displaystyle \underset{-a}{\overset{a}{\int }}\sqrt{\left(1-\frac{x^2}{a^2}\right)}}\\ \text{substitute }x=a\sin \left(u\right)\\ A=2b{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}a{\cos }^2\left(u\right)}du\\ A=2ab{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\cos }^2\left(u\right)}du\\ A=2ab{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{1+\cos \left(2u\right)}{2}}du\\ A=ab\left({\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}1}du+\mathrm{c}{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\cos \left(2u\right)}du\right)\\ A=ab\pi \end{array}$

Therefore,

The integral area formula gives the exact value of area as the simple formula of area of ellipse.