Chapter 11, Problem 53RE

(a)

To determine

To find: The area of $R$ by evaluating an integral in polar coordinates.

Answer to Problem 53RE

The area of $R$ by evaluating an integral in polar coordinates is $\frac{32}{2}$ .

Explanation of Solution

Given:

The equation of the polar curve is defined by $r=\frac{4}{1+\sin \theta }$ for $0\le \theta \le r$ .

Calculation:

The area of the region for the polar curve is given by, $A={\displaystyle \underset{\alpha }{\overset{\beta }{\int }}\frac{1}{2}{r}^{2}d\theta }$ .

Find the area of $R$ by evaluating an integral in polar coordinates.

  $\begin{array}{c}A={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{1}{2}{\left(\frac{4}{1+\sin \theta }\right)}^{2}d\theta }\\ =\frac{32}{2}\end{array}$

Therefore, the area of $R$ by evaluating an integral in polar coordinates is $\frac{32}{2}$ .

(b)

To determine

To convert: The polar equation to rectangular coordinates and proves that the curves are the same.

Answer to Problem 53RE

It is proved that the curve is same.

Explanation of Solution

Given:

The equation of the parabola is $8y=16-x^2$ .

Calculation:

To convert the given polar equation into an equivalent Cartesian equation, use the various mathematical operations and the Polar-Rectangular Cartesian formulas which are,

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

Consider $r=\frac{4}{1+\sin \theta }$ , and simplify.

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

It is known that $r^2=x^2+y^2$ . So,

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

Therefore, it is proved that the curve is same.

(c)

To determine

To set: The integral in rectangular coordinates that gives the area of $R$ .

Answer to Problem 53RE

The integral in rectangular coordinates that gives the area of $R$ is $A={\displaystyle \underset{-4}{\overset{4}{\int }}\left(2-\frac{x^2}{8}\right)dx}$ .

Explanation of Solution

Given:

The equation of the parabola is $8y=16-x^2$ .

Calculation:

The Cartesian form of the curve is,

  $\begin{array}{c}8y=16-x^2\\ y=2-\frac{x^2}{8}\end{array}$

To find the limits substitute $y=0$ in the equation $8y=16-x^2$ .

  $\begin{array}{c}8\left(0\right)=16-{\left(x\right)}^2\\ x^2=16\\ x=\pm 4\end{array}$

So, the integral in rectangular coordinates that gives the area of $R$ is $A={\displaystyle \underset{-4}{\overset{4}{\int }}\left(2-\frac{x^2}{8}\right)dx}$ .

Therefore, the integral in rectangular coordinates that gives the area of $R$ is $A={\displaystyle \underset{-4}{\overset{4}{\int }}\left(2-\frac{x^2}{8}\right)dx}$ .