Skip to content
NCERT Exemplar · Q17

Q.The area of the region bounded by the circle x2+y2=1x^2 + y^2 = 1 is
(A) 2π2\pi sq units
(B) π\pi sq units
(C) 3π3\pi sq units
(D) 4π4\pi sq units

Uttar Pradesh UpmspMCQ· 1mImportance★★★★★
88% · 30/34 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The area of a circle of radius 1 is π⋅12=π\pi \cdot 1^2 = \pi. The region bounded by x2+y2=1x^2 + y^2 = 1 is exactly that circle, so the area is π\pi square units. The correct option is (B).

The question asks for the area of the region bounded by the circle x2+y2=1x^2 + y^2 = 1. This is a standard result from geometry, but let's understand it from the perspective of integration — the "Area Under Curve" approach — so you see why the answer is what it is, not just that it is.

The equation x2+y2=1x^2 + y^2 = 1 describes a circle centered at the origin with radius r=1r = 1. The area of any circle is πr2\pi r^2, so here it's π⋅12=π\pi \cdot 1^2 = \pi. That's the direct answer. But if you were to compute this using calculus, you'd set up an integral for the area between the upper and lower halves of the circle.

  1. Solve for yy in terms of xx.

    From x2+y2=1x^2 + y^2 = 1, we get y=±1−x2y = \pm \sqrt{1 - x^2}. The upper semicircle is y=1−x2y = \sqrt{1 - x^2}, and the lower is y=−1−x2y = -\sqrt{1 - x^2}. The region is symmetric about the xx-axis.

  2. Set up the area integral.

    The area between two curves ytopy_{\text{top}} and ybottomy_{\text{bottom}} from x=ax = a to x=bx = b is ∫ab(ytop−ybottom) dx\int_a^b (y_{\text{top}} - y_{\text{bottom}}) \, dx. Here, ytop=1−x2y_{\text{top}} = \sqrt{1 - x^2}, ybottom=−1−x2y_{\text{bottom}} = -\sqrt{1 - x^2}, and the circle runs from x=−1x = -1 to x=1x = 1. So:

Area=∫−11(1−x2−(−1−x2))dx=∫−1121−x2 dx.\text{Area} = \int_{-1}^{1} \left( \sqrt{1 - x^2} - (-\sqrt{1 - x^2}) \right) dx = \int_{-1}^{1} 2\sqrt{1 - x^2} \, dx.

  1. Evaluate the integral. The integral ∫−111−x2 dx\int_{-1}^{1} \sqrt{1 - x^2} \, dx represents the area of a semicircle of radius 1 (the upper half). That area is π2\frac{\pi}{2}. Multiplying by 2 gives the full circle area: Area=2⋅π2=π.\text{Area} = 2 \cdot \frac{\pi}{2} = \pi. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.