Let's break down your question step by step. ### 1. **Equations to Sketch** 1. \( x = 3y \) 2. \( y = x \) 3. \( y = 2x^2 \) --- ### 2. **Find Intersections** #### **A. \( x = 3y \) and \( y = x \):** Set \( x = 3y \) and \( y = x \): \[ y = x = 3y \implies x = 3x \implies x = 0 \] So \( x = 0, y = 0 \). **Intersection at (0, 0)**. --- #### **B. \( x = 3y \) and \( y = 2x^2 \):** From \( x = 3y \), \( y = \frac{x}{3} \). Plug into \( y = 2x^2 \): \[ \frac{x}{3} = 2x^2 \implies x = 6x^2 \implies 6x^2 - x = 0 \implies x(6x - 1) = 0 \] So \( x = 0 \) or \( x = \frac{1}{6} \). - For \( x = 0 \), \( y = 0 \). - For \( x = \frac{1}{6} \), \( y = 2\left(\frac{1}{6}\right)^2 = 2 \cdot \frac{1}{36} = \frac{1}{18} \). **Intersections at (0, 0) and \(\left(\frac{1}{6}, \frac{1}{18}\right)\)**. --- #### **C. \( y = x \) and \( y = 2x^2 \):** Set \( x = 2x^2 \implies 2x^2 - x = 0 \implies x(2x - 1) = 0 \implies x = 0 \) or \( x = \frac{1}{2} \). - For \( x = 0 \), \( y = 0 \). - For \( x = \frac{1}{2} \), \( y = \frac{1}{2} \). **Intersections at (0, 0) and \(\left(\frac{1}{2}, \frac{1}{2}\right)\)**. --- ### 3. **Summary of Intersection Points** - \( (0, 0) \) - \( \left(\frac{1}{6}, \frac{1}{18}\right) \) - \( \left(\frac{1}{2}, \frac{1}{2}\right) \) --- ### 4. **Sketch and Region** On a sketch: - \( x = 3y \): A line passing through the origin, more vertical than \( y = x \). - \( y = x \): Diagonal line. - \( y = 2x^2 \): An upward opening parabola, vertex at the origin. The region bounded by these three curves is between \( x = 0 \) and \( x = \frac{1}{2} \), bounded below by \( y = 2x^2 \), and above by the smaller of \( y = x \) and \( y = \frac{x}{3} \) (from \( x = 0 \) to \( x = \frac{1}{6} \), \( y = \frac{x}{3} \) is lower than \( y = x \), but actually, for \( x \) between 0 and \( 1/6 \), the region is between the parabola and \( y = \frac{x}{3} \), and for \( x \) from \( 1/6 \) to \( 1/2 \), it's between the parabola and \( y = x \). ### 5. **Set up the Double Integral** - For \( 0 \leq x \leq \frac{1}{6} \): Upper bound is \( y = \frac{x}{3} \) - For \( \frac{1}{6} \leq x \leq \frac{1}{2} \): Upper bound is \( y = x \) - In both, lower bound is \( y = 2x^2 \) **Area:** \[ A = \int_{x=0}^{1/6} \int_{y=2x^2}^{x/3} dy\,dx + \int_{x=1/6}^{1/2} \int_{y=2x^2}^{x} dy\,dx \] Calculate the inner integral: \[ \int_{y=2x^2}^{f(x)} dy = f(x) - 2x^2 \] So, \[ A = \int_{0}^{1/6} \left(\frac{x}{3} - 2x^2\right) dx + \int_{1/6}^{1/2} (x - 2x^2) dx \] --- ### 6. **Compute the Area** #### **First Integral:** \[ \int_{0}^{1/6} \left( \frac{x}{3} - 2x^2 \right) dx \] Integrate: \[ \int \frac{x}{3} dx = \frac{1}{6}x^2 \] \[ \int 2x^2 dx = \frac{2}{3}x^3 \] Evaluate from 0 to \( 1/6 \): \[ = \left[ \frac{1}{6}x^2 - \frac{2}{3}x^3 \right]_{0}^{1/6} \] Plug in \( x = 1/6 \): - \( (1/6)^2 = 1/36 \) - \( (1/6)^3 = 1/216 \) So, \[ \frac{1}{6} \cdot \frac{1}{36} = \frac{1}{216} \] \[ \frac{2}{3} \cdot \frac{1}{216} = \frac{2}{648} = \frac{1}{324} \] So, \[ = \frac{1}{216} - \frac{1}{324} = \frac{3-2}{648} = \frac{1}{648} \] --- #### **Second Integral:** \[ \int_{1/6}^{1/2} (x - 2x^2) dx = \left[ \frac{1}{2}x^2 - \frac{2}{3}x^3 \right]_{1/6}^{1/2} \] Plug in values: For \( x = 1/2 \): - \( (1/2)^2 = 1/4 \), \( (1/2)^3 = 1/8 \) - \( \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \) - \( \frac{2}{3} \cdot \frac{1}{8} = \frac{1}{12} \) So, at \( x = 1/2 \): \[ \frac{1}{8} - \frac{1}{12} = \frac{3-2}{24} = \frac{1}{24} \] Now, for \( x = 1/6 \): - \( (1/6)^2 = 1/36 \), \( (1/6)^3 = 1/216 \) - \( \frac{1}{2} \cdot \frac{1}{36} = \frac{1}{72} \) - \( \frac{2}{3} \cdot \frac{1}{216} = \frac{2}{648} = \frac{1}{324} \) So, \[ \frac{1}{72} - \frac{1}{324} = \frac{9 - 2}{648} = \frac{7}{648} \] So, the second integral is: \[ \frac{1}{24} - \frac{7}{648} \] Let's get common denominators: - \( 1/24 = 27/648 \) - \( 27/648 - 7/648 = 20/648 = 5/162 \) --- ### 7. **Total Area** Add the two results: \[ A = \frac{1}{648} + \frac{5}{162} \] - \( \frac{5}{162} = \frac{20}{648} \) - So, \( \frac{1}{648} + \frac{20}{648} = \frac{21}{648} = \frac{7}{216} \) --- ## **Final Answers** ### **A. Marked Intersections** - \( (0, 0) \) - \( \left(\frac{1}{6}, \frac{1}{18}\right) \) - \( \left(\frac{1}{2}, \frac{1}{2}\right) \) ### **B. Double Integral for Area** \[ A = \int_{0}^{1/6} \left( \frac{x}{3} - 2x^2 \right) dx + \int_{1/6}^{1/2} (x - 2x^2) dx \] ### **C. Area of Shaded Region** \[ \boxed{ A = \frac{7}{216} } \] --- ### **D. Sketch (Description)** - Draw axes and plot the parabola \( y = 2x^2 \) from \( x = 0 \) to \( x = 1/2 \). - Draw \( y = x \) through the origin and \( (1/2, 1/2) \). - Draw \( x = 3y \) (or \( y = \frac{x}{3} \)) through the origin and \( (1/6, 1/18) \). - The region bounded by these is a "curved triangle" shape between these curves, as described above. --- If you need a **hand-drawn sketch** or **further explanation** of the steps, let me know!