Animated Solution for Mathematics - Definite Integration: The area (in square units) of the region enclosed by the ellipse $x^2+3y^2=18$ in the first quadrant below the line $y=x$ is

The region of interest is enclosed by the ellipse $x^2+3y^2=18$ and the line $y=x$ in the first quadrant. To visualize this, imagine a wedge trapped between the origin, the line, and the elliptical arc.

To find the intersection point, we substitute the line equation $y=x$ into the ellipse equation $x^2+3y^2=18$. This yields:

Solving for $x$, we find $x^2=\frac{9}{2}$, which gives the intersection point at $x=\frac{3}{\sqrt{2}}$. This value serves as our critical checkpoint where the boundary of the region transitions.

We define the total area $A$ as the sum of two distinct integrals, $A=A_1+A_2$, because the upper boundary changes at the intersection point.

The first part, $A_1$, represents the area under the line from $x=0$ to $x=\frac{3}{\sqrt{2}}$:

The second part, $A_2$, is the area under the ellipse from $x=\frac{3}{\sqrt{2}}$ to the x-intercept $x=3\sqrt{2}$. Rearranging the ellipse equation gives $y=\frac{1}{\sqrt{3}}\sqrt{18-x^2}$.

The integral is expressed as:

Using the standard integral form $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}(\frac{x}{a})$ with $a=3\sqrt{2}$, we evaluate the bounds. Applying the limits results in:

Simplifying the expression for $A_2$, we obtain:

Finally, we calculate the total area $A$ by summing $A_1$ and $A_2$:

The terms $\frac{9}{4}$ and $-\frac{9}{4}$ cancel out, leaving the final result: