1. Equation of the ellipse

The given ellipse is $x^2+3y^2=18.$

Writing it in standard form: $\frac{x^2}{18}+\frac{y^2}{6}=1.$

So its semi-axes are: $a=3\sqrt{2}, \qquad b=\sqrt{6}.$

Hence total area of the ellipse is $\pi ab=\pi(3\sqrt{2})(\sqrt{6})=6\sqrt{3}\pi.$

Therefore, area of the ellipse in the first quadrant is $\frac{1}{4}\cdot 6\sqrt{3}\pi=\frac{3\sqrt{3}\pi}{2}.$

2. Find where the line $y=x$ meets the ellipse

Substitute $y=x$ into $x^2+3y^2=18.$

Then $x^2+3x^2=18 \implies 4x^2=18 \implies x^2=\frac{9}{2}.$

Since we are in the first quadrant, $x=y=\frac{3}{\sqrt{2}}.$

3. Understand the required region

We need the part of the first-quadrant ellipse that lies below the line $y=x$.

In the first quadrant, the ellipse is $y=\sqrt{\frac{18-x^2}{3}}=\frac{1}{\sqrt{3}}\sqrt{18-x^2}.$

For small $x$, ellipse lies above the line $y=x$; after the intersection point, it lies below it.

So the desired area is:

• from $x=0$ to $x=\frac{3}{\sqrt{2}}$: area between ellipse and line is not included,
• from $x=\frac{3}{\sqrt{2}}$ to $x=3\sqrt{2}$: area under ellipse is included.

A simpler way is:

$\text{Required area} = \text{(first quadrant area of ellipse)} - \text{(part above }y=x\text{ in first quadrant)}.$

That excluded part is bounded by:

• $x=0$ to $x=\frac{3}{\sqrt{2}}$,
• above by ellipse,
• below by line $y=x$.

Thus excluded area is $\int_0^{3/\sqrt{2}} \left(\frac{1}{\sqrt{3}}\sqrt{18-x^2}-x\right)dx.$

So required area is $\frac{3\sqrt{3}\pi}{2}-\int_0^{3/\sqrt{2}} \left(\frac{1}{\sqrt{3}}\sqrt{18-x^2}-x\right)dx.$

This is equivalent to directly computing $\int_{3/\sqrt{2}}^{3\sqrt{2}} \frac{1}{\sqrt{3}}\sqrt{18-x^2}\,dx.$

But there is an even cleaner symmetry argument.

4. Use symmetry with respect to the line $y=x$

The ellipse is not symmetric about $y=x$, so we compute directly.

Let $A=\int_{3/\sqrt{2}}^{3\sqrt{2}} \frac{1}{\sqrt{3}}\sqrt{18-x^2}\,dx.$

Using $\int \sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right),$ with $a=3\sqrt{2}$, we get

$A=\frac{1}{\sqrt{3}}\int_{3/\sqrt{2}}^{3\sqrt{2}} \sqrt{(3\sqrt{2})^2-x^2}\,dx.$

So $A=\frac{1}{\sqrt{3}}\left[\frac{x}{2}\sqrt{18-x^2}+9\sin^{-1}\left(\frac{x}{3\sqrt{2}}\right)\right]_{x=3/\sqrt{2}}^{x=3\sqrt{2}}.$

Now evaluate at the limits.

At $x=3\sqrt{2}$:

$\sqrt{18-18}=0, \qquad \sin^{-1}(1)=\frac{\pi}{2}.$ So value is $9\cdot \frac{\pi}{2}=\frac{9\pi}{2}.$

At $x=\frac{3}{\sqrt{2}}$:

$18-x^2=18-\frac{9}{2}=\frac{27}{2},$ so $\sqrt{18-x^2}=\sqrt{\frac{27}{2}}=\frac{3\sqrt{3}}{\sqrt{2}}.$

Then $\frac{x}{2}\sqrt{18-x^2}=\frac{1}{2}\cdot \frac{3}{\sqrt{2}}\cdot \frac{3\sqrt{3}}{\sqrt{2}}=\frac{9\sqrt{3}}{4}.$

Also, $\frac{x}{3\sqrt{2}}=\frac{3/\sqrt{2}}{3\sqrt{2}}=\frac{1}{2},$ so $\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}.$

Thus lower-limit value is $\frac{9\sqrt{3}}{4}+9\cdot \frac{\pi}{6}=\frac{9\sqrt{3}}{4}+\frac{3\pi}{2}.$

Therefore, $A=\frac{1}{\sqrt{3}}\left(\frac{9\pi}{2}-\frac{9\sqrt{3}}{4}-\frac{3\pi}{2}\right).$

Simplify: