Chapter 11.3, Problem 72E

(a)

To determine

To find: The average value of $r$ with respect to $\theta$ over the following curves $\left(a>0\right)$ for the given cardioid.

Answer to Problem 72E

The average value is $a$ .

Explanation of Solution

Given Data:

The given range is $\alpha \le \theta \le \beta$ .

The expression for the average is $r_{av}=\frac{1}{\beta -\alpha }{\displaystyle {\int }_{\alpha }^{\beta }f\left(\theta \right)d\theta }$ .

The given cardioid is $r=a\left(1-\cos \theta \right)$ .

Calculation:

From the given data the average value of the $r$ is obtained as,

  $\begin{array}{c}{r}_{av}=\frac{1}{2\pi -0}{\displaystyle {\int }_0^{2\pi }a\left(1-\cos \theta \right)d\theta }\\ =\frac{a}{2\pi }{\left[\theta -\sin \theta \right]}_0^{2\pi }\\ =\frac{a}{2\pi }\left(2\pi -\sin 2\pi \right)\\ =a\end{array}$

The average value is $a$ .

(b)

To determine

To find: The average value of $r$ with respect to $\theta$ over the following curves $\left(a>0\right)$ for the given cardioid.

Answer to Problem 72E

The average value is $a$ .

Explanation of Solution

Given Data:

The given range is $\alpha \le \theta \le \beta$ .

The expression for the average is $r_{av}=\frac{1}{\beta -\alpha }{\displaystyle {\int }_{\alpha }^{\beta }f\left(\theta \right)d\theta }$ .

The given cardioid is $r=a$ .

Calculation:

From the given data the average value of the $r$ is obtained as,

  $\begin{array}{c}{r}_{av}=\frac{1}{2\pi -0}{\displaystyle {\int }_0^{2\pi }ad\theta }\\ =\frac{a}{2\pi }{\left[\theta \right]}_0^{2\pi }\\ =\frac{a}{2\pi }\left(2\pi -0\right)\\ =a\end{array}$

The average value is $a$ .

(c)

To determine

To find: The average value of $r$ with respect to $\theta$ over the following curves $\left(a>0\right)$ for the given cardioid.

Answer to Problem 72E

The average value is $a$ .

Explanation of Solution

Given Data:

The given range is $\alpha \le \theta \le \beta$ .

The expression for the average is $r_{av}=\frac{1}{\beta -\alpha }{\displaystyle {\int }_{\alpha }^{\beta }f\left(\theta \right)d\theta }$ .

The given cardioid is $r=a\cos \theta$ .

The region is $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$ .

Calculation:

From the given data the average value of the $r$ is obtained as,

  $\begin{array}{c}{r}_{av}=\frac{1}{\frac{\pi }{2}-\left(-\frac{\pi }{2}\right)}{\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}a\cos \theta d\theta }\\ =\frac{a}{\pi }{\left[\sin \theta \right]}_0^{2\pi }\\ =\frac{a}{2\pi }\left(\sin \frac{\pi }{2}+\sin \frac{\pi }{2}\right)\\ =\frac{a}{\pi }\left(1+1\right)\end{array}$

Solve further as,

  $r_{av}=\frac{2a}{\pi }$

The average value is $\frac{2a}{\pi }$ .