Double Integral Region 92C92C

1. **State the problem:** We need to calculate the double integral $$\iint_R (3xy - x^2 - y^2 + 6) \, dA$$ where the region $R$ is bounded by $x=0$, $y=0$, and $x + 2y = 2$. 2. **Describe the region $R$:** The boundaries are: - $x=0$ (the y-axis), - $y=0$ (the x-axis), - $x + 2y = 2$ (a line). From $x + 2y = 2$, we can express $y$ in terms of $x$: $$y = \frac{2 - x}{2}$$ Since $x \geq 0$ and $y \geq 0$, the limits for $x$ are from $0$ to $2$. 3. **Set up the integral:** We integrate first with respect to $y$ from $0$ to $\frac{2 - x}{2}$, then with respect to $x$ from $0$ to $2$: $$\int_0^2 \int_0^{\frac{2 - x}{2}} (3xy - x^2 - y^2 + 6) \, dy \, dx$$ 4. **Integrate with respect to $y$:** Calculate the inner integral: $$\int_0^{\frac{2 - x}{2}} (3xy - x^2 - y^2 + 6) \, dy = \left[ \frac{3x y^2}{2} - x^2 y - \frac{y^3}{3} + 6y \right]_0^{\frac{2 - x}{2}}$$ Substitute $y = \frac{2 - x}{2}$: $$= \frac{3x}{2} \left( \frac{2 - x}{2} \right)^2 - x^2 \left( \frac{2 - x}{2} \right) - \frac{1}{3} \left( \frac{2 - x}{2} \right)^3 + 6 \left( \frac{2 - x}{2} \right)$$ Simplify each term: - $\left( \frac{2 - x}{2} \right)^2 = \frac{(2 - x)^2}{4}$ - $\left( \frac{2 - x}{2} \right)^3 = \frac{(2 - x)^3}{8}$ So inner integral becomes: $$\frac{3x}{2} \cdot \frac{(2 - x)^2}{4} - x^2 \cdot \frac{2 - x}{2} - \frac{1}{3} \cdot \frac{(2 - x)^3}{8} + 6 \cdot \frac{2 - x}{2}$$ Simplify coefficients: $$= \frac{3x (2 - x)^2}{8} - \frac{x^2 (2 - x)}{2} - \frac{(2 - x)^3}{24} + 3(2 - x)$$ 5. **Expand and simplify:** - Expand $(2 - x)^2 = 4 - 4x + x^2$ - Expand $(2 - x)^3 = (2 - x)(4 - 4x + x^2) = 8 - 12x + 6x^2 - x^3$ Substitute: $$\frac{3x (4 - 4x + x^2)}{8} - \frac{x^2 (2 - x)}{2} - \frac{8 - 12x + 6x^2 - x^3}{24} + 3(2 - x)$$ Calculate each term: - $\frac{3x (4 - 4x + x^2)}{8} = \frac{3x \cdot 4}{8} - \frac{3x \cdot 4x}{8} + \frac{3x \cdot x^2}{8} = \frac{12x}{8} - \frac{12x^2}{8} + \frac{3x^3}{8} = \frac{3x}{2} - \frac{3x^2}{2} + \frac{3x^3}{8}$ - $- \frac{x^2 (2 - x)}{2} = - \frac{2x^2 - x^3}{2} = - x^2 + \frac{x^3}{2}$ - $- \frac{8 - 12x + 6x^2 - x^3}{24} = - \frac{8}{24} + \frac{12x}{24} - \frac{6x^2}{24} + \frac{x^3}{24} = - \frac{1}{3} + \frac{x}{2} - \frac{x^2}{4} + \frac{x^3}{24}$ - $3(2 - x) = 6 - 3x$ 6. **Combine all terms:** $$\left( \frac{3x}{2} - \frac{3x^2}{2} + \frac{3x^3}{8} \right) + \left( - x^2 + \frac{x^3}{2} \right) + \left( - \frac{1}{3} + \frac{x}{2} - \frac{x^2}{4} + \frac{x^3}{24} \right) + (6 - 3x)$$ Group like terms: - $x^3$ terms: $\frac{3x^3}{8} + \frac{x^3}{2} + \frac{x^3}{24} = \frac{9x^3}{24} + \frac{12x^3}{24} + \frac{x^3}{24} = \frac{22x^3}{24} = \frac{11x^3}{12}$ - $x^2$ terms: $- \frac{3x^2}{2} - x^2 - \frac{x^2}{4} = - \frac{6x^2}{4} - \frac{4x^2}{4} - \frac{x^2}{4} = - \frac{11x^2}{4}$ - $x$ terms: $\frac{3x}{2} + \frac{x}{2} - 3x = \frac{4x}{2} - 3x = 2x - 3x = -x$ - Constants: $- \frac{1}{3} + 6 = \frac{18}{3} - \frac{1}{3} = \frac{17}{3}$ So the inner integral simplifies to: $$\frac{11x^3}{12} - \frac{11x^2}{4} - x + \frac{17}{3}$$ 7. **Integrate with respect to $x$ from $0$ to $2$:** $$\int_0^2 \left( \frac{11x^3}{12} - \frac{11x^2}{4} - x + \frac{17}{3} \right) dx = \left[ \frac{11x^4}{48} - \frac{11x^3}{12} - \frac{x^2}{2} + \frac{17x}{3} \right]_0^2$$ 8. **Evaluate at $x=2$:** - $\frac{11 \cdot 2^4}{48} = \frac{11 \cdot 16}{48} = \frac{176}{48} = \frac{11}{3}$ - $- \frac{11 \cdot 2^3}{12} = - \frac{11 \cdot 8}{12} = - \frac{88}{12} = - \frac{22}{3}$ - $- \frac{2^2}{2} = - \frac{4}{2} = -2$ - $\frac{17 \cdot 2}{3} = \frac{34}{3}$ Sum these: $$\frac{11}{3} - \frac{22}{3} - 2 + \frac{34}{3} = \left( \frac{11 - 22 + 34}{3} \right) - 2 = \frac{23}{3} - 2 = \frac{23}{3} - \frac{6}{3} = \frac{17}{3}$$ 9. **Evaluate at $x=0$:** All terms are zero. 10. **Final answer:** $$\boxed{\frac{17}{3}}$$ This is the value of the double integral over the region $R$.