CUET UG 2025 — Mathematics Calculus

Given curve: $y = \sqrt{16-x^2}$y=16−x2​

$$y^2 = 16-x^2$$y2=16−x2

$$x^2 + y^2 = 16$$x2+y2=16

This is the equation of a circle with center at origin $(0, 0)$(0,0) and radius $= \sqrt{16} = 4$=16​=4 units.

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The original equation is $y = \sqrt{16-x^2}$y=16−x2​ with the square root.

Since square root gives only positive values, $y \geq 0$y≥0

This represents only the upper half of the circle (above the x-axis).

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When $y = 0$y=0:

$$0 = \sqrt{16-x^2}$$0=16−x2​

$$0 = 16-x^2$$0=16−x2

$$x^2 = 16$$x2=16

$$x = \pm 4$$x=±4

The curve touches the x-axis at $x = -4$x=−4 and $x = 4$x=4.

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The region bounded by the curve and x-axis is a semicircle (upper half).

Area of semicircle $= \frac{1}{2} \times \pi r^2$=21​×πr2

Where radius $r = 4$r=4:

Area $= \frac{1}{2} \times \pi \times (4)^2$=21​×π×(4)2

Area $= \frac{1}{2} \times \pi \times 16$=21​×π×16

Area $= 8\pi$=8π sq. units

Therefore, the area of the region is $8\pi$8π square units.