Q. 23 - Calculus

Question

Find a definite integral expression that represents the area of the given region in polar plane and then find the exact value of the expression 

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

Step-by-Step Solution

Answer

The integral can be given as $A=2{\int }_0^{\frac{\sqrt{3}}{2}}(\sqrt{1-x^2}-x)dy$

And the value of the integral is $0.634$

Step 1: Given information

We are given a circle $x^2+y^2=1 and right of the vertical line x=\frac{1}{2}$

Step 2: Find the value of values of y

We have,

$$\begin{gathered} x^2+y^2=1 \\ {y}^2=1-\frac{1}{4} \\ {y}^2=\frac{3}{4} \\ y=\pm \frac{3}{4} \end{gathered}$$

Now the integral can be given as

$A=2{\int }_0^{\frac{\sqrt{3}}{2}}(\sqrt{1-x^2}-x)dy$

Now the area can be given as

$$\begin{gathered} A=2(\sqrt{1-x^2}-x){\int }_0^{\frac{\sqrt{3}}{2}}dy \\ A=(\sqrt{1-x^2}-x)\left(\sqrt{3}\right) \end{gathered}$$

Step 3: Find the exact value

We substitute $x=\frac{1}{2}$ in the equation we get

$$\begin{gathered} A=(\sqrt{1-\frac{1}{4}}-\frac{1}{2})\left(\sqrt{3}\right) \\ A=(\frac{\sqrt{3}}{2}-\frac{1}{2})\left(\sqrt{3}\right) \\ A=\frac{3-\sqrt{3}}{2} \\ A=0.634 \end{gathered}$$