Q.Area of the region in the first quadrant enclosed by the x-axis, the line $y = x$ and the circle $x^2 + y^2 = 32$ is
(A) $16\pi$ sq units
(B) $4\pi$ sq units
(C) $32\pi$ sq units
(D) $24$ sq units

Concept: Area Under Curve — the required region is bounded by the x-axis, the line $y = x$, and the circle $x^2 + y^2 = 32$ in the first quadrant.

Step 1: Find intersection points.

The line $y = x$ meets the circle $x^2 + x^2 = 32 \Rightarrow 2x^2 = 32 \Rightarrow x^2 = 16 \Rightarrow x = 4$ (first quadrant). So the intersection is $(4,4)$. The circle meets the x-axis at $x = \sqrt{32} = 4\sqrt{2}$.

Step 2: Set up the area.

From $x = 0$ to $x = 4$, the upper curve is $y = x$. From $x = 4$ to $x = 4\sqrt{2}$, the upper curve is $y = \sqrt{32 - x^2}$ (the circle).

Area $= \int_0^4 x\,dx + \int_4^{4\sqrt{2}} \sqrt{32 - x^2}\,dx$.

Step 3: Evaluate.

First integral: $\left[\frac{x^2}{2}\right]_0^4 = 8$.

Second integral: use formula $\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$ with $a = 4\sqrt{2}$. …

The problem asks for the area in the first quadrant bounded by three curves: the x‑axis ($y=0$), the line $y=x$, and the circle $x^2+y^2=32$. The key is to see that the circle’s centre is at the origin, and the line $y=x$ makes a $45^\circ$ angle with the x‑axis. So the region is simply a circular sector — no integration needed if you recognise this.

Let’s walk through it.

1. Understand the circle.

   The equation $x^2+y^2=32$ gives radius $r = \sqrt{32} = 4\sqrt{2}$. The full circle area is $\pi r^2 = \pi \cdot 32 = 32\pi$ sq units.

2. Identify the boundaries in the first quadrant.

   • The x‑axis ($y=0$) is the lower boundary.
   • The line $y=x$ passes through the origin at $45^\circ$ to the x‑axis.
   • The circle is the outer boundary. All three meet at the origin? Actually, the x‑axis and $y=x$ meet at $(0,0)$, but the circle does not pass through the origin — it passes through $(4\sqrt{2},0)$ on the x‑axis and through $(4,4)$ where $y=x$ meets the circle (since $x^2+x^2=2x^2=32 \Rightarrow x^2=16 \Rightarrow x=4$ in the first quadrant). So the region is not a triangle; it’s the part of the circle between the ray $y=0$ and the ray $y=x$, from the origin out to the circle.

3. Recognise the sector.

   The region is exactly the sector of the circle bounded by the two radii: one along the positive x‑axis (angle $0$) and one along the line $y=x$ (angle $45^\circ = \pi/4$). The arc of the circle from $(4\sqrt{2},0)$ to $(4,4)$ completes the boundary. So the area is simply the area of a circular sector with radius $4\sqrt{2}$ and central angle $\pi/4$.

4. Compute the sector area. Here $\theta = \pi/4$, $r = 4\sqrt{2}$. …

Method: First-quadrant area bounded by an axis, a ray through the origin, and a circle

When the boundaries are the $x$-axis, a line through the origin such as $y = x$, and a circle centred at the origin, the region is a circular sector — though it can also be found by splitting the area integral at the line–circle intersection.

Steps

Step 1: Find where the ray meets the circle.

Substitute the line into the circle equation to get the intersection point; this is where the top boundary switches from the line to the circular arc.

Step 2 (sector shortcut): use the angle.

A ray $y = x$ makes $\frac{\pi}{4}$ with the $x$-axis, so the region is a sector of angle $\theta$ and radius $r$ (from $x^2 + y^2 = r^2$):

Step 3 (integration route): split the x-range. …

Common Mistakes

Mistake 1: Taking the radius as $32$ instead of $\sqrt{32}=4\sqrt2$.

Why it's wrong: $32$ is $r^2$; using it as $r$ makes every area far too large. Correct approach: the circle has $r=4\sqrt2$, so the full area is $\pi(4\sqrt2)^2=32\pi$ and the required region is only a fraction of it.

Mistake 2: Reporting the whole circle area $32\pi$ (or a quadrant).

Why it's wrong: the region is the sector between the ray $y=0$ and the ray $y=x$, a central angle of just $\tfrac{\pi}{4}$ — one-eighth of the circle. Correct approach: use the sector area $\tfrac12 r^2\theta=\tfrac12(32)\tfrac{\pi}{4}=4\pi$. …