Find area enclosed by circle x²+y²=4 using integration

easy CBSE JEE-MAIN • 3 min read
Tags Area Under Curves Intuition

Question

Find the area enclosed by the circle x2+y2=4x^2 + y^2 = 4 using integration.

Solution — Step by Step

The circle x2+y2=4x^2 + y^2 = 4 has centre at origin and radius r=2r = 2 (since r2=4r^2 = 4).

We know the area of a circle is πr2=4π\pi r^2 = 4\pi. Integration will confirm this.

Strategy: Integrate to find the area of the upper semicircle (above the x-axis), then double it for the full area. Alternatively, use the first-quadrant quarter and multiply by 4.

From x2+y2=4x^2 + y^2 = 4: y=4−x2y = \sqrt{4 - x^2} (positive square root = upper semicircle)

The full circle spans x∈[−2,2]x \in [-2, 2].

Area of upper semicircle = ∫−224−x2 dx\int_{-2}^{2} \sqrt{4 - x^2}\, dx

We use the formula: ∫abr2−x2 dx\int_a^b \sqrt{r^2 - x^2}\, dx = area under the curve of upper semicircle.

Specifically: ∫−rrr2−x2 dx=πr22\int_{-r}^{r} \sqrt{r^2 - x^2}\, dx = \frac{\pi r^2}{2}

Here r=2r = 2, so:

∫−224−x2 dx=π×42=2π\int_{-2}^{2} \sqrt{4 - x^2}\, dx = \frac{\pi \times 4}{2} = 2\pi

Area of full circle = 2 × (area of upper semicircle)

Area=2×2π=4π\text{Area} = 2 \times 2\pi = \mathbf{4\pi}

This confirms the standard formula πr2=π(2)2=4π\pi r^2 = \pi(2)^2 = 4\pi.

Why This Works

The integral ∫r2−x2 dx\int \sqrt{r^2 - x^2}\, dx evaluates the “height” of the upper semicircle at each x, and integrating sums up all these thin vertical strips. This is the geometric definition of area under a curve.

For those who want the full derivation using trigonometric substitution:

Let x=2sin⁡θx = 2\sin\theta, so dx=2cos⁡θ dθdx = 2\cos\theta\, d\theta.

When x=−2x = -2: θ=−π/2\theta = -\pi/2; when x=2x = 2: θ=π/2\theta = \pi/2.

4−x2=4−4sin⁡2θ=2cos⁡θ\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = 2\cos\theta

∫−π/2π/22cos⁡θ⋅2cos⁡θ dθ=4∫−π/2π/2cos⁡2θ dθ\int_{-\pi/2}^{\pi/2} 2\cos\theta \cdot 2\cos\theta\, d\theta = 4\int_{-\pi/2}^{\pi/2} \cos^2\theta\, d\theta =4∫−π/2π/21+cos⁡2θ2 dθ=2[θ+sin⁡2θ2]−π/2π/2=2[π/2+0−(−π/2)−0]=2π= 4\int_{-\pi/2}^{\pi/2} \frac{1 + \cos 2\theta}{2}\, d\theta = 2\left[\theta + \frac{\sin 2\theta}{2}\right]_{-\pi/2}^{\pi/2} = 2[\pi/2 + 0 - (-\pi/2) - 0] = 2\pi

Total area = 2×2π=4π2 \times 2\pi = 4\pi. ✓

Alternative Method

Using the quarter-circle (first quadrant, x∈[0,2]x \in [0, 2], y≥0y \geq 0):

Area=4∫024−x2 dx=4×π=4π\text{Area} = 4\int_0^2 \sqrt{4 - x^2}\, dx = 4 \times \pi = 4\pi

Since ∫024−x2 dx=14×π(2)2=π\int_0^2 \sqrt{4-x^2}\, dx = \frac{1}{4} \times \pi(2)^2 = \pi (area of quarter circle of radius 2).

For CBSE Class 12 and JEE integration problems, the standard result ∫0rr2−x2 dx=πr24\int_0^r \sqrt{r^2 - x^2}\, dx = \frac{\pi r^2}{4} (quarter circle area) is worth memorizing directly. It saves time in longer problems where circle area appears as one part of a larger question. Always specify that y=r2−x2y = \sqrt{r^2 - x^2} is the upper semicircle (not lower) when setting up the integral.

Common Mistake

Students sometimes try to compute the area directly as ∫−22y dx\int_{-2}^{2} y\, dx without specifying which half of the circle they’re integrating. The circle is not a function — for each xx (except ±2\pm 2), there are two yy values: +4−x2+\sqrt{4-x^2} (upper) and −4−x2-\sqrt{4-x^2} (lower). If you integrate from x=−2x = -2 to x=2x = 2, you must use only the positive (upper) or negative (lower) half. Another error: trying to compute ∫−22(yupper−ylower) dx=∫−2224−x2 dx\int_{-2}^{2} (y_{upper} - y_{lower})\, dx = \int_{-2}^{2} 2\sqrt{4-x^2}\, dx is the correct full approach but students sometimes forget the factor of 2 when combining.

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