The area (in sq. units) of the bigger portion of region enclosed by the curves $4x^2 + 9y^2 = 36$ and $2x + 3y = 6$ is

The given curves are:

$4x^2 + 9y^2 = 36 \implies \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$

So $a = 3$ and $b = 2$.

$2x + 3y = 6 \implies \dfrac{x}{3} + \dfrac{y}{2} = 1$

The intercept form tells us the line passes through $(3, 0)$ and $(0, 2)$. Both points satisfy the ellipse equation, so the line meets the ellipse at exactly these two points and cuts it into two parts.

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Total area of the ellipse:

$A_{\text{ellipse}} = \pi a b = \pi(3)(2) = 6\pi$

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The smaller portion lies in the first quadrant between the ellipse (above) and the line (below).

$A_{\text{small}} = \displaystyle\int_0^3 \left[\frac{2}{3}\sqrt{9 - x^2} - \frac{6 - 2x}{3}\right] dx$

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Evaluating the ellipse part:

$\displaystyle\int_0^3 \frac{2}{3}\sqrt{9 - x^2}\, dx = \frac{2}{3} \int_0^3 \sqrt{9 - x^2}\, dx$

Using $\displaystyle\int_0^a \sqrt{a^2 - x^2}\, dx = \frac{\pi a^2}{4}$:

$= \dfrac{2}{3} \times \dfrac{9\pi}{4}$

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

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Evaluating the line part:

$\displaystyle\int_0^3 \frac{6 - 2x}{3}\, dx = \frac{1}{3}\Big[6x - x^2\Big]_0^3$

$= \dfrac{1}{3}(18 - 9)$

$= 3$

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$A_{\text{small}} = \dfrac{3\pi}{2} - 3$

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$A_{\text{big}} = A_{\text{ellipse}} - A_{\text{small}}$

$= 6\pi - \left(\dfrac{3\pi}{2} - 3\right)$

$= 6\pi - \dfrac{3\pi}{2} + 3$

$= \dfrac{12\pi - 3\pi}{2} + 3$

$= \dfrac{9\pi}{2} + 3$

$= \dfrac{9\pi + 6}{2}$

$= \dfrac{3}{2}(3\pi + 2)$