7. Evaluate the double integral \( \iiint_{D} y d A \) where \( D \) is the triangular region with vertices \( (0,0),(1,1) \), and \( (4,0) \)

We wish to evaluate \[ \iint_{D} y\, dA, \] where the triangular region \( D \) is defined by the vertices \((0,0)\), \((1,1)\), and \((4,0)\). **Step 1. Find the equations of the boundary lines.** - **Between \((0,0)\) and \((1,1)\):** The line has slope \[ m = \frac{1-0}{1-0} = 1, \] so its equation is \[ y = x. \] - **Between \((1,1)\) and \((4,0)\):** The slope is \[ m = \frac{0-1}{4-1} = -\frac{1}{3}. \] Using the point-slope form with point \((1,1)\), \[ y - 1 = -\frac{1}{3}(x-1), \] which simplifies to \[ y = 1 - \frac{x-1}{3} = \frac{4}{3} - \frac{x}{3}. \] - **Between \((4,0)\) and \((0,0)\):** This is the horizontal line \( y = 0 \). **Step 2. Describe the region \( D \) using integration limits.** We can project the region onto the \(x\)-axis. Notice: - \( x \) varies from \( 0 \) to \( 4 \). - For a fixed \( x \), \( y \) varies from the line \( y = 0 \) (bottom) to a line that depends on \( x \): - For \( x \) between \( 0 \) and \( 1 \), the upper boundary is the line \( y = x \). - For \( x \) between \( 1 \) and \( 4 \), the upper boundary is the line \( y = \frac{4}{3} - \frac{x}{3} \). Thus, we split the integral into two regions: \[ \iint_{D} y\, dA = \int_{x=0}^{1} \int_{y=0}^{x} y\, dy\, dx + \int_{x=1}^{4} \int_{y=0}^{\frac{4}{3} - \frac{x}{3}} y\, dy\, dx. \] **Step 3. Evaluate the inner integrals.** - **For \(0 \le x \le 1\):** \[ \int_{y=0}^{x} y\, dy = \left.\frac{1}{2} y^2 \right|_{0}^{x} = \frac{1}{2} x^2. \] - **For \(1 \le x \le 4\):** \[ \int_{y=0}^{\frac{4}{3} - \frac{x}{3}} y\, dy = \left.\frac{1}{2} y^2 \right|_{0}^{\frac{4}{3} - \frac{x}{3}} = \frac{1}{2} \left(\frac{4}{3} - \frac{x}{3}\right)^2. \] **Step 4. Set up and compute the outer integrals.** - **First Integral:** \[ I_1 = \int_{x=0}^{1} \frac{1}{2} x^2\, dx = \frac{1}{2} \int_{0}^{1} x^2\, dx. \] We have \[ \int_{0}^{1} x^2\, dx = \left.\frac{x^3}{3}\right|_{0}^{1} = \frac{1}{3}, \] so \[ I_1 = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}. \] - **Second Integral:** \[ I_2 = \int_{x=1}^{4} \frac{1}{2} \left(\frac{4}{3} - \frac{x}{3}\right)^2 dx. \] Factor the constant: \[ I_2 = \frac{1}{2} \int_{x=1}^{4} \left(\frac{4-x}{3}\right)^2 dx = \frac{1}{2} \cdot \frac{1}{9} \int_{x=1}^{4} (4-x)^2\, dx = \frac{1}{18} \int_{x=1}^{4} (4-x)^2\, dx. \] Make a substitution: let \( u = 4-x \) so that \( du = - dx \). When \( x=1 \), \( u = 3 \); when \( x=4 \), \( u = 0 \). Then, \[ I_2 = \frac{1}{18} \int_{u=3}^{0} u^2\, (-du) = \frac{1}{18} \int_{u=0}^{3} u^2\, du. \] Evaluate the integral: \[ \int_{0}^{3} u^2\, du = \left.\frac{u^3}{3}\right|_{0}^{3} = \frac{27}{3} = 9, \] so \[ I_2 = \frac{1}{18} \times 9 = \frac{1}{2}. \] **Step 5. Combine the results.** The value of the integral is: \[ \iint_{D} y\, dA = I_1 + I_2 = \frac{1}{6} + \frac{1}{2} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}. \] Thus, the value of the double integral is \[ \boxed{\frac{2}{3}}. \]