Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x² + y² = 32 is

The correct option is 4π sq. units.

Let us solve this problem step-by-step to find the area of the region in the first quadrant enclosed by the x-axis, the line $y=x$, and the circle $x^2+y^2=32$.

Step 1: Identify the given equations and region boundaries
1. The equation of the circle is:

$x^2+y^2=32$

This is a circle centered at the origin $(0,0)$ with radius:

$r=\sqrt{32}=4\sqrt{2}$

2. The boundary line is:

$y=x$

3. The other boundaries are the x-axis ($y=0$) and the condition for the first quadrant where both $x\ge 0$ and $y\ge 0$.

Step 2: Find the point of intersection
To find where the line $y=x$ intersects the circle, we substitute $y=x$ into the circle's equation:

$x^2+x^2=32$

$2x^2=32$

$x^2=16$

Since we are looking for the intersection in the first quadrant, we take the positive square root:

$x=4$

Since $y=x$, the intersection point is $(4,4)$.

Step 3: Calculate the area of the region
We can find the area using two different methods: geometric sector analysis (highly intuitive) and integration.

Method A: Geometric Analysis (Sector of a Circle)
The line $y=x$ bisects the first quadrant, making an angle of $\theta =\frac{\pi }{4}$ radians (or $45°$) with the positive x-axis.
The region lies between the x-axis ($\theta =0$) and the line $y=x$ ($\theta =\frac{\pi }{4}$) inside the circle of radius $r=\sqrt{32}$.
Therefore, the region is simply a sector of the circle with a central angle of $\theta =\frac{\pi }{4}$.
The formula for the area of a sector is:

$\text{Area}=\frac{1}{2}{r}^2\theta$

Substituting the values of $r^2=32$ and $\theta =\frac{\pi }{4}$:

$\text{Area}=\frac{1}{2}\times 32\times \frac{\pi }{4}=16\times \frac{\pi }{4}=4\pi \text{sq. units}$

Method B: Using Integration
We split the region into two parts along the vertical line $x=4$:
1. Part 1: From $x=0$ to $x=4$, bounded above by the line $y=x$.
2. Part 2: From $x=4$ to $x=4\sqrt{2}$, bounded above by the circle $y=\sqrt{32-x^2}$.
The total area $A$ is given by:

$A={\int }_0^{4}xdx+{\int }_4^{4\sqrt{2}}\sqrt{32-x^2}dx$

Evaluating the first integral:

${\int }_0^{4}xdx={\left[\frac{x^2}{2}\right]}_0^4=\frac{16}{2}-0=8$

Evaluating the second integral using the standard formula $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)$ with $a^2=32$:

${\int }_4^{4\sqrt{2}}\sqrt{32-x^2}dx={[\frac{x}{2}}_4^{4\sqrt{2}}$

Substitute the upper limit $x=4\sqrt{2}$:

$\frac{4\sqrt{2}}{2}\sqrt{32-32}+16{\sin }^{-1}\left(1\right)=0+16\left(\frac{\pi }{2}\right)=8\pi$

Substitute the lower limit $x=4$:

$\frac{4}{2}\sqrt{32-16}+16{\sin }^{-1}\left(\frac{4}{4\sqrt{2}}\right)=2\sqrt{16}+16{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)=2\left(4\right)+16\left(\frac{\pi }{4}\right)=8+4\pi$

Subtracting the lower limit value from the upper limit value:

${\int }_4^{4\sqrt{2}}\sqrt{32-x^2}dx=8\pi -(8+4\pi )=4\pi -8$

Adding both parts to find the total area:

$A=8+(4\pi -8)=4\pi \text{sq. units}$