Question

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: 

• A. \( \sqrt{3\pi} + \frac{3}{4} \)
• B. \( \sqrt{3}\pi \)
• C. \( \sqrt{3\pi} - \frac{3}{4} \)
• D. \( \sqrt{3\pi} + 1 \)

Answer 1

The Correct Option is B

We start with the given expression for the area:

\[ \text{Area} = \int_{0}^{\frac{3}{\sqrt{2}}} x \, dx + \int_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}} \frac{\sqrt{18 - x^2}}{3} \, dx \]

Evaluating the first integral:

\[ \int_{0}^{\frac{3}{\sqrt{2}}} x \, dx = \frac{1}{2} \left( x^2 \right)_{0}^{\frac{3}{\sqrt{2}}} = \frac{1}{2} \left( \frac{9}{2} \right) = \frac{9}{4} \]

Now, for the second integral:

\[ \int_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}} \frac{\sqrt{18 - x^2}}{3} \, dx = \frac{1}{3} \int_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}} \sqrt{18 - x^2} \, dx \]

We know the standard integral formula:

\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]

Substituting \( a = 3\sqrt{2} \):

\[ \int_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}} \sqrt{18 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{18 - x^2} + 9 \sin^{-1}\left( \frac{x}{3\sqrt{2}} \right) \right]_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}} \]

Thus,

\[ \text{Area} = \frac{1}{2}\left(\frac{9}{2}\right) + \frac{1}{\sqrt{3}} \left[ 9\sin^{-1}(1) - \frac{3}{2\sqrt{2}} \cdot 3\sqrt{3} - 9\sin^{-1}\left(\frac{1}{2}\right) \right] \]

Evaluating each term:

\[ \text{Area} = \frac{9}{4} + \frac{1}{\sqrt{3}}\left( \frac{9\pi}{2} - \frac{9\sqrt{3}}{4} - \frac{9\pi}{6} \right) \]

Simplifying further:

\[ \text{Area} = \sqrt{3}\pi \]

Final Answer:

\[ \boxed{\sqrt{3}\pi} \]

Answer 2

Given the equation of the ellipse: The given ellipse equation is:

\[ \frac{x^2}{18} + \frac{y^2}{6} = 1. \]

This is an ellipse centered at the origin with semi-major axis \(\sqrt{18} = 3\sqrt{2}\) along the \(x\)-axis and semi-minor axis \(\sqrt{6}\) along the \(y\)-axis.

The line \(y = x\) intersects the ellipse in the first quadrant. Substituting \(y = x\) into the ellipse equation:

\[ \frac{x^2}{18} + \frac{x^2}{6} = 1 \implies \frac{4x^2}{18} = 1 \implies x^2 = \frac{9}{2}. \]

Thus, the point of intersection is:

\[ \left( x, y \right) = \left( \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} \right). \]

Step 1: Area under the ellipse below \(y = x\)
The area of the elliptical segment in the first quadrant below the line \(y = x\) is given by:

\[ \int_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}} \sqrt{\frac{18 - x^2}{3}} \, dx. \]

Substituting the limits and evaluating (from the given solution in the image):

\[ \frac{1}{\sqrt{3}} \left[ \frac{18 - x^2}{2} + \frac{18}{2} \sin^{-1} \left( \frac{x}{3\sqrt{2}} \right) \right]_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}}. \]

This simplifies to:

\[ \frac{1}{\sqrt{3}} \left[ 9 \cdot \frac{\pi}{2} - \frac{3}{2\sqrt{2}} \cdot \sqrt{3\sqrt{2}} - 9 \cdot \frac{\pi}{6} \right]. \]

Step 2: Total Area Calculation
From the triangle and ellipse calculations, the required area is:

Required Area = \(\sqrt{3}\pi\).

So, Final Answer: \(\sqrt{3}\pi\)