The area (in sq. units) of the region in the first quadrant bounded by $y = 3\sqrt{1-x^2}$, $x \in [0,1]$ and the x-axis is equal to

The area bounded by the curve, the x-axis, and $x \in [0,1]$ is:

$A = \int_0^1 3\sqrt{1-x^2}\;dx$

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Let $x = \sin\theta \implies dx = \cos\theta\;d\theta$

When $x = 0$, $\theta = 0$

When $x = 1$, $\theta = \frac{\pi}{2}$

$A = 3\int_0^{\pi/2} \sqrt{1 - \sin^2\theta}\;\cos\theta\;d\theta$

Since $1 - \sin^2\theta = \cos^2\theta$:

$A = 3\int_0^{\pi/2} \cos^2\theta\;d\theta$

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Using $\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}$:

$A = \dfrac{3}{2}\int_0^{\pi/2} (1 + \cos 2\theta)\;d\theta$

$= \dfrac{3}{2}\left[\theta + \dfrac{\sin 2\theta}{2}\right]_0^{\pi/2}$

$= \dfrac{3}{2}\left[\left(\dfrac{\pi}{2} + \dfrac{\sin \pi}{2}\right) - \left(0 + 0\right)\right]$

$= \dfrac{3}{2} \times \dfrac{\pi}{2}$

$= \dfrac{3\pi}{4}$

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The curve $y = 3\sqrt{1-x^2}$ represents the upper half of the ellipse $\dfrac{x^2}{1} + \dfrac{y^2}{9} = 1$ with $a=1$, $b=3$.

Full ellipse area $= \pi ab = 3\pi$

First quadrant portion $= \dfrac{1}{4} \times 3\pi = \dfrac{3\pi}{4}$ sq. units