1. **State the problem:** Find the area between the ellipse given by $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ and the line $$x + y = 2$$. 2. **Rewrite the line equation:** Solve for $$y$$ in terms of $$x$$: $$y = 2 - x$$. 3. **Express the ellipse equation for $$y$$:** $$\frac{x^2}{9} + \frac{y^2}{4} = 1 \implies \frac{y^2}{4} = 1 - \frac{x^2}{9} \implies y^2 = 4\left(1 - \frac{x^2}{9}\right) = 4 - \frac{4x^2}{9}$$ So, $$y = \pm \sqrt{4 - \frac{4x^2}{9}} = \pm \frac{2}{3} \sqrt{9 - x^2}$$. 4. **Find intersection points:** Solve the system: $$x + y = 2$$ and $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$. Substitute $$y = 2 - x$$ into ellipse: $$\frac{x^2}{9} + \frac{(2 - x)^2}{4} = 1$$ Multiply both sides by 36 (LCM of 9 and 4): $$4x^2 + 9(2 - x)^2 = 36$$ Expand: $$4x^2 + 9(4 - 4x + x^2) = 36$$ $$4x^2 + 36 - 36x + 9x^2 = 36$$ Combine like terms: $$13x^2 - 36x + 36 = 36$$ Subtract 36: $$13x^2 - 36x = 0$$ Factor: $$x(13x - 36) = 0$$ So, $$x = 0$$ or $$x = \frac{36}{13}$$. Find corresponding $$y$$ values: - For $$x=0$$, $$y=2-0=2$$. - For $$x=\frac{36}{13}$$, $$y=2 - \frac{36}{13} = \frac{26}{13} - \frac{36}{13} = -\frac{10}{13}$$. 5. **Set up the integral for the area:** The area between the curves from $$x=0$$ to $$x=\frac{36}{13}$$ is $$\text{Area} = \int_0^{\frac{36}{13}} \left[(2 - x) - \frac{2}{3} \sqrt{9 - x^2}\right] dx$$ Here, $$2 - x$$ is the line (upper curve) and $$\frac{2}{3} \sqrt{9 - x^2}$$ is the upper half of the ellipse. 6. **Calculate the integral:** Split the integral: $$\int_0^{\frac{36}{13}} (2 - x) dx - \int_0^{\frac{36}{13}} \frac{2}{3} \sqrt{9 - x^2} dx$$ First integral: $$\int_0^{\frac{36}{13}} (2 - x) dx = \left[2x - \frac{x^2}{2}\right]_0^{\frac{36}{13}} = 2 \cdot \frac{36}{13} - \frac{(\frac{36}{13})^2}{2} = \frac{72}{13} - \frac{1296}{169 \cdot 2} = \frac{72}{13} - \frac{1296}{338}$$ Convert $$\frac{72}{13}$$ to denominator 338: $$\frac{72}{13} = \frac{72 \times 26}{13 \times 26} = \frac{1872}{338}$$ So, $$\frac{1872}{338} - \frac{1296}{338} = \frac{576}{338} = \frac{288}{169}$$ Second integral: $$\int_0^{\frac{36}{13}} \frac{2}{3} \sqrt{9 - x^2} dx = \frac{2}{3} \int_0^{\frac{36}{13}} \sqrt{9 - x^2} dx$$ Use formula: $$\int \sqrt{a^2 - x^2} dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \arcsin\left(\frac{x}{a}\right) + C$$ Here, $$a=3$$. Evaluate: $$\int_0^{\frac{36}{13}} \sqrt{9 - x^2} dx = \left[\frac{x}{2} \sqrt{9 - x^2} + \frac{9}{2} \arcsin\left(\frac{x}{3}\right)\right]_0^{\frac{36}{13}}$$ Calculate each term at $$x=\frac{36}{13}$$: - $$\sqrt{9 - \left(\frac{36}{13}\right)^2} = \sqrt{9 - \frac{1296}{169}} = \sqrt{\frac{1521}{169} - \frac{1296}{169}} = \sqrt{\frac{225}{169}} = \frac{15}{13}$$ - So, $$\frac{x}{2} \sqrt{9 - x^2} = \frac{36}{13} \cdot \frac{15}{13} \cdot \frac{1}{2} = \frac{36 \times 15}{13 \times 13 \times 2} = \frac{540}{338} = \frac{270}{169}$$ - $$\arcsin\left(\frac{36}{13 \times 3}\right) = \arcsin\left(\frac{36}{39}\right) = \arcsin\left(\frac{12}{13}\right)$$ Therefore, $$\int_0^{\frac{36}{13}} \sqrt{9 - x^2} dx = \frac{270}{169} + \frac{9}{2} \arcsin\left(\frac{12}{13}\right) - 0 = \frac{270}{169} + \frac{9}{2} \arcsin\left(\frac{12}{13}\right)$$ Multiply by $$\frac{2}{3}$$: $$\frac{2}{3} \times \left(\frac{270}{169} + \frac{9}{2} \arcsin\left(\frac{12}{13}\right)\right) = \frac{540}{507} + 3 \arcsin\left(\frac{12}{13}\right) = \frac{180}{169} + 3 \arcsin\left(\frac{12}{13}\right)$$ 7. **Final area:** $$\text{Area} = \frac{288}{169} - \left(\frac{180}{169} + 3 \arcsin\left(\frac{12}{13}\right)\right) = \frac{108}{169} - 3 \arcsin\left(\frac{12}{13}\right)$$ This is the exact area between the ellipse and the line. **Answer:** $$\boxed{\text{Area} = \frac{108}{169} - 3 \arcsin\left(\frac{12}{13}\right)}$$