Area Enclosed

1. **State the problem:** Find the area enclosed by the graphs of the circle $x^2 + y^2 = 4$, the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$, and the line $x + y = 2$. 2. **Analyze each curve:** - Circle: radius 2 centered at origin. - Ellipse: semi-major axis 3 along x-axis, semi-minor axis 2 along y-axis. - Line: intercepts at $(2,0)$ and $(0,2)$. 3. **Find intersection points:** - Between circle and line: Substitute $y = 2 - x$ into $x^2 + y^2 = 4$: $$x^2 + (2 - x)^2 = 4$$ $$x^2 + 4 - 4x + x^2 = 4$$ $$2x^2 - 4x + 4 = 4$$ $$2x^2 - 4x = 0$$ $$2x(x - 2) = 0$$ So $x=0$ or $x=2$. Corresponding $y$ values: $y=2$ or $y=0$. Intersection points: $(0,2)$ and $(2,0)$. - Between ellipse and line: Substitute $y = 2 - x$ into ellipse equation: $$\frac{x^2}{9} + \frac{(2 - x)^2}{4} = 1$$ Multiply both sides by 36 (LCM of 9 and 4): $$4x^2 + 9(2 - x)^2 = 36$$ $$4x^2 + 9(4 - 4x + x^2) = 36$$ $$4x^2 + 36 - 36x + 9x^2 = 36$$ $$13x^2 - 36x + 36 = 36$$ $$13x^2 - 36x = 0$$ $$x(13x - 36) = 0$$ So $x=0$ or $x=\frac{36}{13}$. Corresponding $y$ values: $y=2$ or $y=2 - \frac{36}{13} = \frac{26}{13} - \frac{36}{13} = -\frac{10}{13}$. Intersection points: $(0,2)$ and $(\frac{36}{13}, -\frac{10}{13})$. - Between circle and ellipse: Solve system: $$x^2 + y^2 = 4$$ $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ Multiply ellipse equation by 36: $$4x^2 + 9y^2 = 36$$ From circle: $y^2 = 4 - x^2$. Substitute into ellipse: $$4x^2 + 9(4 - x^2) = 36$$ $$4x^2 + 36 - 9x^2 = 36$$ $$-5x^2 + 36 = 36$$ $$-5x^2 = 0$$ $$x=0$$ Then $y^2 = 4$, so $y=\pm 2$. Intersection points: $(0,2)$ and $(0,-2)$. 4. **Identify enclosed region:** The three curves intersect at $(0,2)$, $(2,0)$, and $(\frac{36}{13}, -\frac{10}{13})$ (ellipse-line), and $(0,-2)$ (circle-ellipse). The region bounded by all three is the area inside the circle and ellipse and below the line between $(0,2)$ and $(2,0)$. 5. **Set up integrals:** - From $x=0$ to $x=2$, the line is $y=2 - x$. - The circle upper half: $y=\sqrt{4 - x^2}$. - The ellipse lower half: $y=-2\sqrt{1 - \frac{x^2}{9}}$. 6. **Determine integration intervals:** - Between $x=0$ and $x=\frac{36}{13}$, the ellipse is below the line. - Between $x=\frac{36}{13}$ and $x=2$, the circle is below the line. 7. **Calculate area:** Area = area under line minus area under ellipse (from 0 to $\frac{36}{13}$) plus area under circle minus area under line (from $\frac{36}{13}$ to 2): $$\text{Area} = \int_0^{\frac{36}{13}} (2 - x - (-2\sqrt{1 - \frac{x^2}{9}})) dx + \int_{\frac{36}{13}}^2 (\sqrt{4 - x^2} - (2 - x)) dx$$ Simplify integrands: $$= \int_0^{\frac{36}{13}} (2 - x + 2\sqrt{1 - \frac{x^2}{9}}) dx + \int_{\frac{36}{13}}^2 (\sqrt{4 - x^2} - 2 + x) dx$$ 8. **Evaluate integrals:** - First integral: $$I_1 = \int_0^{\frac{36}{13}} (2 - x) dx + 2 \int_0^{\frac{36}{13}} \sqrt{1 - \frac{x^2}{9}} dx$$ - Second integral: $$I_2 = \int_{\frac{36}{13}}^2 \sqrt{4 - x^2} dx - \int_{\frac{36}{13}}^2 (2 - x) dx$$ 9. **Compute each part:** - $\int (2 - x) dx = 2x - \frac{x^2}{2}$ - $\int \sqrt{1 - \frac{x^2}{9}} dx$ can be computed by substitution $x=3\sin\theta$: $$\int \sqrt{1 - \sin^2\theta} \cdot 3\cos\theta d\theta = 3 \int \cos^2\theta d\theta$$ Use identity $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$. - $\int \sqrt{4 - x^2} dx$ is the area under a circle segment: $$\int \sqrt{4 - x^2} dx = \frac{x}{2} \sqrt{4 - x^2} + 2 \arcsin\left(\frac{x}{2}\right) + C$$ - $\int (2 - x) dx$ as above. 10. **Calculate numeric values:** - $I_1$: $$\int_0^{\frac{36}{13}} (2 - x) dx = \left[2x - \frac{x^2}{2}\right]_0^{\frac{36}{13}} = 2 \cdot \frac{36}{13} - \frac{(36/13)^2}{2} = \frac{72}{13} - \frac{1296}{169 \cdot 2} = \frac{72}{13} - \frac{1296}{338}$$ Approximate: $$\frac{72}{13} \approx 5.5385, \quad \frac{1296}{338} \approx 3.8343$$ So $5.5385 - 3.8343 = 1.7042$. - For $2 \int_0^{\frac{36}{13}} \sqrt{1 - \frac{x^2}{9}} dx$: Substitute $x=3\sin\theta$, when $x=0$, $\theta=0$, when $x=\frac{36}{13}$, $\sin\theta = \frac{36}{39} = \frac{12}{13}$, so $\theta = \arcsin(12/13)$. $$2 \cdot 3 \int_0^{\arcsin(12/13)} \cos^2\theta d\theta = 6 \int_0^{\arcsin(12/13)} \frac{1 + \cos 2\theta}{2} d\theta = 3 \left[ \theta + \frac{\sin 2\theta}{2} \right]_0^{\arcsin(12/13)}$$ Calculate: $$\theta = \arcsin(12/13) \approx 1.176$$ $$\sin 2\theta = 2 \sin\theta \cos\theta = 2 \cdot \frac{12}{13} \cdot \frac{5}{13} = \frac{120}{169} \approx 0.710$$ So: $$3 (1.176 + \frac{0.710}{2}) = 3 (1.176 + 0.355) = 3 (1.531) = 4.593$$ - Therefore, $I_1 = 1.7042 + 4.593 = 6.2972$. - $I_2$: $$\int_{\frac{36}{13}}^2 \sqrt{4 - x^2} dx = \left[ \frac{x}{2} \sqrt{4 - x^2} + 2 \arcsin\left(\frac{x}{2}\right) \right]_{\frac{36}{13}}^2$$ At $x=2$: $$\frac{2}{2} \cdot 0 + 2 \cdot \arcsin(1) = 0 + 2 \cdot \frac{\pi}{2} = \pi \approx 3.1416$$ At $x=\frac{36}{13} \approx 2.769$, but since $2.769 > 2$, this is invalid; re-check: $\frac{36}{13} \approx 2.769$ is incorrect, actually $\frac{36}{13} = 2.769$ which is greater than 2, so the previous assumption is wrong. Re-examine step 6: $\frac{36}{13} \approx 2.769$ which is greater than 2, so the ellipse-line intersection is outside the circle's domain. So the line intersects the ellipse at $x=0$ and $x=2$ (since $x=2$ is the circle-line intersection). Therefore, the region is bounded between $x=0$ and $x=2$. The area is the region bounded by the line and circle from $x=0$ to $x=2$, and the ellipse from $x=0$ to $x=0$ (no area). So the enclosed area is the area between the circle and line from $x=0$ to $x=2$. Calculate: $$\text{Area} = \int_0^2 (\sqrt{4 - x^2} - (2 - x)) dx$$ - Compute: $$\int_0^2 \sqrt{4 - x^2} dx = \left[ \frac{x}{2} \sqrt{4 - x^2} + 2 \arcsin\left(\frac{x}{2}\right) \right]_0^2 = \pi$$ $$\int_0^2 (2 - x) dx = \left[ 2x - \frac{x^2}{2} \right]_0^2 = 4 - 2 = 2$$ - So area: $$\pi - 2 \approx 3.1416 - 2 = 1.1416$$ **Final answer:** $$\boxed{1.1416}$$ This is the area enclosed by the circle $x^2 + y^2 = 4$ and the line $x + y = 2$ within the ellipse's domain.