Q.Show that the area of the smaller region bounded by the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ and the line $\dfrac{x}{2}+\dfrac{y}{2}=1$ is $\dfrac{3}{2}(\pi-2)$ sq. units.

Area of an Ellipse

A circle of radius $r$ has area $\pi r^2$. An ellipse is a circle that has been stretched by different amounts along two perpendicular directions, so it is natural to expect its area to be a stretched version of $\pi r^2$. The standard ellipse centred at the origin is

where $a$ is the semi-major (or semi-minor) axis along $x$ and $b$ is the semi-axis along $y$. The result we want is beautifully simple:

Notice that when $a = b = r$ the ellipse becomes a circle and $\pi ab$ collapses to $\pi r^2$ — a good sanity check.

Finding it by integration

Because the ellipse is symmetric about both axes, we compute the area of the piece in the first quadrant and multiply by $4$. Solving the equation for the upper half gives

As $x$ runs from $0$ to $a$ this traces the first-quadrant arc, so

The integral $\displaystyle\int_0^a \sqrt{a^2 - x^2}\, dx$ is the area of a quarter-circle of radius $a$, which equals $\dfrac{\pi a^2}{4}$. (You may also get it from the standard result $\int \sqrt{a^2 - x^2}\,dx = \tfrac{x}{2}\sqrt{a^2-x^2} + \tfrac{a^2}{2}\sin^{-1}\tfrac{x}{a} + C$.) Therefore

The intuition

The factor $\dfrac{b}{a}$ in front is exactly the vertical stretch that turns a circle of radius $a$ into this ellipse: it scales every height by $b/a$, and scaling all heights scales the area by the same ratio. Multiplying the circle's area $\pi a^2$ by $b/a$ gives $\pi ab$. …