Let's break it down step by step: ## **Step 1: Write the equations clearly** Given: - \( y = \frac{1}{3}x \) - \( y = x \) - \( y = 2x^2 \) --- ## **Step 2: Find the points of intersection** ### **A. Intersection of \( y = x \) and \( y = 2x^2 \):** Set \( x = 2x^2 \): \[ x = 2x^2 \implies 2x^2 - x = 0 \implies x(2x - 1) = 0 \] So, \[ x = 0 \quad \text{or} \quad x = \frac{1}{2} \] Corresponding \( y \): - For \( x = 0 \), \( y = 0 \) - For \( x = \frac{1}{2} \), \( y = \frac{1}{2} \) **Points:** \((0,0)\) and \(\left(\frac{1}{2}, \frac{1}{2}\right)\) --- ### **B. Intersection of \( y = \frac{1}{3}x \) and \( y = 2x^2 \):** Set \( \frac{1}{3}x = 2x^2 \): \[ \frac{1}{3}x = 2x^2 \implies 2x^2 - \frac{1}{3}x = 0 \implies x(2x - \frac{1}{3}) = 0 \] So, \[ x = 0 \quad \text{or} \quad x = \frac{1}{6} \] Corresponding \( y \): - For \( x = 0 \), \( y = 0 \) - For \( x = \frac{1}{6} \), \( y = \frac{1}{3}\cdot\frac{1}{6} = \frac{1}{18} \) **Points:** \((0,0)\) and \(\left(\frac{1}{6}, \frac{1}{18}\right)\) --- ### **C. Intersection of \( y = x \) and \( y = \frac{1}{3}x \):** Set \( x = \frac{1}{3}x \implies x - \frac{1}{3}x = 0 \implies \frac{2}{3}x = 0 \implies x = 0 \) So, \( y = 0 \) **Point:** \((0, 0)\) --- #### **Summary of Intersections:** - \((0, 0)\) : all three curves intersect - \(\left(\frac{1}{2}, \frac{1}{2}\right)\) : \(y=x\) and \(y=2x^2\) - \(\left(\frac{1}{6}, \frac{1}{18}\right)\) : \(y=\frac{1}{3}x\) and \(y=2x^2\) --- ## **Step 3: Sketch and Identify the Region** - \(y=2x^2\) is a parabola opening upwards. - \(y=x\) is a straight line through the origin at \(45^\circ\). - \(y=\frac{1}{3}x\) is a straight line through the origin but less steep. - All three intersect at \((0,0)\). - The three curves form a bounded region between \(x=0\) and \(x=\frac{1}{2}\). **Between \(x=0\) and \(x=\frac{1}{6}\):** - Top: \(y=\frac{1}{3}x\) - Bottom: \(y=2x^2\) **Between \(x=\frac{1}{6}\) and \(x=\frac{1}{2}\):** - Top: \(y=x\) - Bottom: \(y=2x^2\) **See sketch:** ``` y ^ | / | / y = x | / | / | / | / y = (1/3)x | / | / |/______________> x \ / \ / V y = 2x^2 ``` The bounded region is between \(x=0\) and \(x=\frac{1}{2}\), above \(y=2x^2\), below \(y=\frac{1}{3}x\) for \(x\in[0, 1/6]\), and below \(y=x\) for \(x\in[1/6, 1/2]\). --- ## **Step 4: Set Up the Double Integral** **Area**: \[ A = \int_{x=0}^{1/6} \left[ \frac{1}{3}x - 2x^2 \right] dx + \int_{x=1/6}^{1/2} \left[ x - 2x^2 \right] dx \] --- ## **Step 5: Compute the Area** ### **A. First Integral (\(x=0\) to \(x=1/6\))** \[ I_1 = \int_{0}^{1/6} \left( \frac{1}{3}x - 2x^2 \right) dx \] \[ = \int_{0}^{1/6} \frac{1}{3}x dx - \int_{0}^{1/6} 2x^2 dx \] \[ = \frac{1}{3} \int_{0}^{1/6} x dx - 2 \int_{0}^{1/6} x^2 dx \] \[ = \frac{1}{3} \left[ \frac{1}{2}x^2 \right]_{0}^{1/6} - 2 \left[ \frac{1}{3}x^3 \right]_{0}^{1/6} \] \[ = \frac{1}{3} \cdot \frac{1}{2} \left( \left(\frac{1}{6}\right)^2 - 0 \right) - 2 \cdot \frac{1}{3} \left( \left(\frac{1}{6}\right)^3 - 0 \right) \] \[ = \frac{1}{6} \cdot \frac{1}{36} - \frac{2}{3} \cdot \frac{1}{216} \] \[ = \frac{1}{216} - \frac{2}{648} \] \[ = \frac{1}{216} - \frac{1}{324} \] Find common denominator: \[ \frac{1}{216} = \frac{3}{648}, \quad \frac{1}{324} = \frac{2}{648} \] So, \[ = \frac{3}{648} - \frac{2}{648} = \frac{1}{648} \] --- ### **B. Second Integral (\(x=1/6\) to \(x=1/2\))** \[ I_2 = \int_{1/6}^{1/2} (x - 2x^2) dx \] \[ = \int_{1/6}^{1/2} x dx - \int_{1/6}^{1/2} 2x^2 dx \] \[ = \left[ \frac{1}{2}x^2 \right]_{1/6}^{1/2} - 2\left[ \frac{1}{3}x^3 \right]_{1/6}^{1/2} \] \[ = \frac{1}{2} \left( \left(\frac{1}{2}\right)^2 - \left(\frac{1}{6}\right)^2 \right) - \frac{2}{3} \left( \left(\frac{1}{2}\right)^3 - \left(\frac{1}{6}\right)^3 \right) \] \[ = \frac{1}{2} \left( \frac{1}{4} - \frac{1}{36} \right) - \frac{2}{3} \left( \frac{1}{8} - \frac{1}{216} \right) \] \[ = \frac{1}{2} \left( \frac{9}{36} - \frac{1}{36} \right) - \frac{2}{3} \left( \frac{27}{216} - \frac{1}{216} \right) \] \[ = \frac{1}{2} \left( \frac{8}{36} \right) - \frac{2}{3} \left( \frac{26}{216} \right) \] \[ = \frac{4}{36} - \frac{2}{3} \cdot \frac{13}{108} \] \[ = \frac{1}{9} - \frac{26}{324} \] \[ = \frac{36}{324} - \frac{26}{324} \] \[ = \frac{10}{324} = \frac{5}{162} \] --- ### **C. Total Area** \[ A = I_1 + I_2 = \frac{1}{648} + \frac{5}{162} \] Convert \(\frac{5}{162}\) to denominator 648: \[ \frac{5}{162} = \frac{5 \times 4}{162 \times 4} = \frac{20}{648} \] So, \[ A = \frac{1}{648} + \frac{20}{648} = \frac{21}{648} = \frac{7}{216} \] --- ## **Step 6: Final Answer** ### **The area of the required region is:** \[ \boxed{\frac{7}{216}} \] ### **Intersections:** - \((0,0)\) - \(\left(\frac{1}{6}, \frac{1}{18}\right)\) - \(\left(\frac{1}{2}, \frac{1}{2}\right)\) **Shaded region**: Bounded by \(y=2x^2\) below, \(y=\frac{1}{3}x\) above for \(x\in[0,1/6]\), and \(y=x\) above for \(x\in[1/6,1/2]\). --- ### **Summary Table:** | Intersection | x-value | y-value | |-----------------------------|--------------|-----------------| | All three curves | \(0\) | \(0\) | | \(y=\frac{1}{3}x\), \(2x^2\)| \(\frac{1}{6}\)| \(\frac{1}{18}\)| | \(y=x\), \(2x^2\) | \(\frac{1}{2}\)| \(\frac{1}{2}\) | --- ### **Area:** \[ \boxed{\frac{7}{216}} \] If you want a decimal: \[ \frac{7}{216} \approx 0.0324 \] --- **Let me know if you need a diagram or further explanation!**