Chapter 10, Problem 53EP

i.

To determine

To calculate: To solve the area of $R$ as an integral in polar coordinates.

Answer to Problem 53EP

The Area of an integral is $8{\displaystyle {\int }_0^{\pi }{\left(\frac{1}{1+\sin \theta }\right)}^{2}d\theta }$ .

Explanation of Solution

Given information: The given curve as follows:

The region $R$ in the $xy-$ plane is bounded below by the $x-$ axis and above by the polar curve defined by $r=\frac{4}{1+\sin \theta }$ for $0\le \theta \le \pi$ .

Calculation:

Since, the area can be calculated as:

  $\text{Area=}{\displaystyle {\int }_a^b{r}^{2}d\theta }$

The area of $R$ as an integral.

The area in an integral form can be represented as follows:

  $\begin{array}{l}\text{Area}=\frac{1}{2}{\displaystyle {\int }_0^{\pi }{\left(\frac{4}{1+\sin \theta }\right)}^{2}d\theta }\\ \text{Area}=\frac{1}{2}{\displaystyle {\int }_0^{\pi }\left(\frac{16}{{\left(1+\sin \theta \right)}^2}\right)d\theta }\\ \text{Area}=8{\displaystyle {\int }_0^{\pi }{\left(\frac{1}{1+\sin \theta }\right)}^{2}d\theta }\end{array}$ .

ii.

To determine

To calculate: To solve the polar equation to rectangular coordinates and the curves should be same.

Answer to Problem 53EP

The conversion into rectangular system is $8y=-x^2+16$ and both curves are same.

Explanation of Solution

Given information:

The given curve as follows:

The region $R$ in the $xy-$ plane is bounded below by the $x-$ axis and above by the polar curve defined by $r=\frac{4}{1+\sin \theta }$ for $0\le \theta \le \pi$ . The curve resembles an arch of the parabola $8y=16-x^2$ .

Calculation:

Since, the curves can be calculated as:

  $\begin{array}{c}r=\frac{4}{1+\sin \theta }\\ r\left(1+\sin \theta \right)=4\\ r+r\sin \theta =4\\ r=4-r\sin \theta \end{array}$

The polar coordinates are:

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \\ r^2=x^2+y^2\end{array}$

These values can be substituted in the above equations as follows:

  $\begin{array}{c}r=4-r\sin \theta \\ r^2={\left(4-r\sin \theta \right)}^2\\ r^2=16+r^2{\sin }^2\theta -8r\sin \theta \\ x^2+y^2=16+y^2-8y\\ x^2=16-8y\\ 8y=-x^2+16\end{array}$

Hence, both curves are same.

iii.

To determine

To calculate: To set up the area of $R$ .

Answer to Problem 53EP

The setup of an integral $={\displaystyle {\int }_{-4}^4\left(2-\frac{x^2}{8}\right)}dx$ .

Explanation of Solution

Given information:

The given curve as follows:

The region $R$ in the $xy-$ plane is bounded below by the $x-$ axis and above by the polar curve defined by $r=\frac{4}{1+\sin \theta }$ for $0\le \theta \le \pi$ .

To set up the integral of $R$ in rectangular coordinates form.

Calculation:

Since, the area can be calculated as:

  $\text{Area}={\displaystyle {\int }_a^{b}ydx}$

The area of $R$ as an integral.

The value of $y$ varies from $0to\frac{16-x^2}{8}$ .

  $\begin{array}{c}16-x^2=0\\ x^2=16\\ x=\pm 4\end{array}$

Here, it can be solved as:

  $\begin{array}{c}\text{Area}={\displaystyle {\int }_{-4}^4\frac{16-x^2}{8}}-0dx\\ ={\displaystyle {\int }_{-4}^4\left(2-\frac{x^2}{8}\right)}dx\end{array}$