Q. 35 - Calculus

Question

In Exercises 29–38, find an iterated integral in polar coordinates that represents the area of the given region in the polar plane and then evaluate the integral. 

The region inside the circle $x^2 + y^2 = 1$ and to the right of the vertical line $x = \frac{1}{2}$ .

Step-by-Step Solution

Answer

The needed region's area is $A=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$

Step 1 : Given Information

The equation is $x^2 + y^2 = 1$   and   line $x=\frac{1}{2}$

Step 2: Simplification

The goal of this issue is to calculate the size of the region that lies inside the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=1$ and to the right of the vertical line $\text{x=}\frac{1}{2}$

Draw a circle ${\mathrm{x}}^2+{\mathrm{y}}^2=1$ and a line on a piece of paper on $\text{x=}\frac{1}{2}$

Graph of  ${\mathrm{x}}^2+{\mathrm{y}}^2=1\text{ and }\mathrm{x}=\frac{1}{2}$

Step 3: Finding the required region's size

Area of the circle's inner region $x^2+y^2=1$ and the right-hand side of the vertical line $x=\frac{1}{2}$ is ABCD.

The required area can be expressed as 

$A={\int }_{x=1/2}^{x=1} {\int }_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}} dxdy$

Around the x- axis, the required region is symmetric. 

That is, 

$A=2{\int }_{x=1/2}^{x=1} {\int }_{y=0}^{y=\sqrt{1-x^2}} dxdy$

Integrate first with respect to the y,$\mathrm{A}={\int }_{\mathrm{x}=1/2}^{\mathrm{x}=1} {\int }_{\mathrm{y}=-\sqrt{1-{\mathrm{x}}^2}}^{\mathrm{y}=\sqrt{1-{\mathrm{x}}^2}} \mathrm{dxdy}$

$\begin{array}{r}\mathrm{A}=2{\int }_{\mathrm{x}=1/2}^{\mathrm{x}=1} [\mathrm{y}{]}_0^{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\\ \mathrm{A}=2{\int }_{\mathrm{x}=1/2}^{\mathrm{x}=1} \sqrt{1-{\mathrm{x}}^2}\mathrm{dx}\\ \mathrm{A}=2{\left[\frac{\mathrm{x}}{2}\sqrt{1-{\mathrm{x}}^2}+\frac{1}{2}{\sin }^{-1}\mathrm{x}\right]}_{1/2}^1\end{array}$

Set the boundaries

$\begin{array}{r}\mathrm{A}=2\left[\left\{\frac{1}{2}{\sin }^{-1}1\right\}-\left\{\frac{1/2}{2}\sqrt{1-(1/2{)}^2}+\frac{1}{2}{\sin }^{-1}(1/2)\right\}\right]\\ \mathrm{A}=2\left[\left\{\frac{1}{2},\frac{\mathrm{\pi }}{2}\right\}-\left\{\frac{1}{4},\frac{\sqrt{3}}{2}+\frac{1}{2}\cdot \frac{\mathrm{\pi }}{6}\right\}\right]\\ \mathrm{A}=2\left[\frac{\mathrm{\pi }}{6}-\frac{\sqrt{3}}{8}\right]\\ \mathrm{A}=\frac{\mathrm{\pi }}{3}-\frac{\sqrt{3}}{4}\end{array}$

As a result, the needed region's area is $A=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$