
How do you find the area inside one loop of the lemniscate ${{r}^{2}}=5\sin 2\theta $?
Answer
542.4k+ views
Hint: We explain the number of ways position of a point or equation can be expressed in different forms. We also explain the ways the representation works for polar and cartesian form. Then we convert the given equation into rectangular form using the relations $x=r\cos \theta ;y=r\sin \theta $. We find the limits of the curve and through integration find the area.
Complete step by step answer:
The given equation ${{r}^{2}}=5\sin 2\theta $ is a representation of the polar form. r represents the distance and $\theta $ represents the angle.
We need to convert the given equation ${{r}^{2}}=5\sin 2\theta $ into the rectangular form.
From multiple angle theorem we get $\sin 2\theta =2\sin \theta \cos \theta $.
The relation between these two forms in two-dimensional is
$x=r\cos \theta ;y=r\sin \theta ;{{x}^{2}}+{{y}^{2}}={{r}^{2}}$.
From the relations we get $\sin \theta =\dfrac{y}{r},\cos \theta =\dfrac{x}{r}$.
We now replace the values in the equation ${{r}^{2}}=5\sin 2\theta $ to get
\[\begin{align}
& {{r}^{2}}=5\sin 2\theta \\
& \Rightarrow {{r}^{2}}=10\sin \theta \cos \theta \\
& \Rightarrow {{r}^{2}}=10\times \dfrac{y}{r}\times \dfrac{x}{r} \\
& \Rightarrow {{r}^{4}}=10xy \\
\end{align}\]
We now replace the value of ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ for the equation \[{{r}^{4}}=10xy\]. The revised equation becomes \[{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=10xy\]. The equation is an equation of loop.
The loop goes from $0\le \theta \le \dfrac{\pi }{2}$. We integrate the equation for the area A.
So, $A=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{5\sin 2\theta d\theta }=\dfrac{-5}{4}\left[ \cos 2\theta \right]_{0}^{\dfrac{\pi }{2}}=\dfrac{-5}{4}\left( -2 \right)=\dfrac{5}{2}$.
Therefore, the area inside one loop of the lemniscate ${{r}^{2}}=5\sin 2\theta $ is $\dfrac{5}{2}$ unit.
Note: There are always two ways to represent any point, equation in our general 2-D and 3-D surfaces. One being the polar form and other one being the cartesian form. In case of polar form, we use the distance and the angle from the origin to get the position of the point or curve. In case of rectangular form, we use the coordinates from the origin to get the position of the point or curve. For two dimensional things we have X-Y and we take the perpendicular distances from the axes.
Complete step by step answer:
The given equation ${{r}^{2}}=5\sin 2\theta $ is a representation of the polar form. r represents the distance and $\theta $ represents the angle.
We need to convert the given equation ${{r}^{2}}=5\sin 2\theta $ into the rectangular form.
From multiple angle theorem we get $\sin 2\theta =2\sin \theta \cos \theta $.
The relation between these two forms in two-dimensional is
$x=r\cos \theta ;y=r\sin \theta ;{{x}^{2}}+{{y}^{2}}={{r}^{2}}$.
From the relations we get $\sin \theta =\dfrac{y}{r},\cos \theta =\dfrac{x}{r}$.
We now replace the values in the equation ${{r}^{2}}=5\sin 2\theta $ to get
\[\begin{align}
& {{r}^{2}}=5\sin 2\theta \\
& \Rightarrow {{r}^{2}}=10\sin \theta \cos \theta \\
& \Rightarrow {{r}^{2}}=10\times \dfrac{y}{r}\times \dfrac{x}{r} \\
& \Rightarrow {{r}^{4}}=10xy \\
\end{align}\]
We now replace the value of ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ for the equation \[{{r}^{4}}=10xy\]. The revised equation becomes \[{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=10xy\]. The equation is an equation of loop.
The loop goes from $0\le \theta \le \dfrac{\pi }{2}$. We integrate the equation for the area A.
So, $A=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{5\sin 2\theta d\theta }=\dfrac{-5}{4}\left[ \cos 2\theta \right]_{0}^{\dfrac{\pi }{2}}=\dfrac{-5}{4}\left( -2 \right)=\dfrac{5}{2}$.
Therefore, the area inside one loop of the lemniscate ${{r}^{2}}=5\sin 2\theta $ is $\dfrac{5}{2}$ unit.
Note: There are always two ways to represent any point, equation in our general 2-D and 3-D surfaces. One being the polar form and other one being the cartesian form. In case of polar form, we use the distance and the angle from the origin to get the position of the point or curve. In case of rectangular form, we use the coordinates from the origin to get the position of the point or curve. For two dimensional things we have X-Y and we take the perpendicular distances from the axes.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

