Chapter 10, Problem 5RWDT

To determine

To find: Why does the integral formula using polar coordinates for the area enclosed by $r={\left(\sin \theta \right)}^4$ evaluated from 0 to $2\pi$ give the right answer ,when the same formula for $r={\left(\sin \theta \right)}^5$ on the same interval gives twice the right answer?

Answer to Problem 5RWDT

The curve $r={\left(\sin \theta \right)}^4$ has the double area of the curve $r={\left(\sin \theta \right)}^5$ because it has symmetry about polar axis .

Explanation of Solution

Given information:

The equation of polar curves is given as:

  $r={\left(\sin \theta \right)}^4$ and $r={\left(\sin \theta \right)}^5$ .

The area of $r={\left(\sin \theta \right)}^5$ is double the area of $r={\left(\sin \theta \right)}^4$

Calculation:

The reason behind the area of $r={\left(\sin \theta \right)}^5$ being double the area of $r={\left(\sin \theta \right)}^4$ can be understood by the shape of both of curves.

Both curves represents the basic polar curve $r=a\sin \theta$ which is a circle with diameter $a$ and center at $\left(\frac{a}{2},\theta \right)$ .

Substitute $\theta =-\theta$ in both curves:

  $\begin{array}{l}r={\left(\sin \left(-\theta \right)\right)}^5,r={\left(\sin \left(-\theta \right)\right)}^4\\ r={\left(-\sin \theta \right)}^5,r={\left(-\sin \theta \right)}^4\\ r=-{\left(\sin \theta \right)}^5,r={\left(\sin \theta \right)}^4\end{array}$

As displayed above the equation of the curve remains the same if $\theta =-\theta$ this gives the idea that the curve has symmetry about polar axis but the curve $r={\left(\sin \theta \right)}^5$ is converted into $r=-{\left(\sin \theta \right)}^5$ therefore it doesn’t have symmetry about the polar axis.

The shape of the curve $r={\left(\sin \theta \right)}^5$ is given below:

  

The curve $r={\left(\sin \theta \right)}^5$ passes through $\left(0,1\right)$

The shape of the curve $r={\left(\sin \theta \right)}^4$ is given below:

  

The curve $r={\left(\sin \theta \right)}^5$ passes through $\left(0,1\right)$ and $\left(0,-1\right)$

It is clear from the above figures that the curve $r={\left(\sin \theta \right)}^4$ has symmetry about $x$ axis .

Therefore whatever the area is above the $x$ axis it will be equal below the $x$ axis as well.

This gives the explanation to the fact that the curve $r={\left(\sin \theta \right)}^4$ has the double the area of the curve $r={\left(\sin \theta \right)}^5$ .