## Step 1: Set the Integral for the Mass The of the plate is found by integrating the density over the region \( R \): \[ M = \iint_R \rho(x, y) \, dA \] Given: \[ \rho(x, y) = \frac{y^2}{x} \] The region \( R \) is bounded by: - \( x = 6 \) - \( y = \frac{x}{3} \) - \( xy = 3 \implies y = \frac{3}{x} \) For a fixed \( x \) (from the left curve to the right line): - The lower bound: \( y = \frac{3}{x} \) - The upper bound: \( y = \frac{x}{3} \) - \( x \) runs from the intersection of \( y = \frac{x}{3} \) and \( y = \frac{3}{x} \) to \( x = 6 \). Let's find the intersection: \[ \frac{x}{3} = \frac{3}{x} \implies x^2 = 9 \implies x = 3 \quad (\text{since } x > ) \] So, \( x \) runs from 3 to 6. Set up the double integral: \[ M = \int_{x=3}^{6} \int_{y = \frac{3}{x}}^{\frac{x}{3}} \frac{y^2}{x} \, dy\,dx \] **Explanation**: We identified the region of integration and set up the bounds for \( x \) and \( y \). The density function is included in the integrand. --- ## Step 2: Evaluate the Integral First, integrate with respect to \( y \): \[ \int_{y = \frac{3}{x}}^{\frac{x}{3}} \frac{y^2}{x} dy = \frac{1}{x} \int_{y = \frac{3}{x}}^{\frac{x}{3}} y^2 dy \] \[ = \frac{1}{x} \left[ \frac{y^3}{3} \right]_{y = \frac{3}{x}}^{y = \frac{x}{3}} \] \[ = \frac{1}{x} \left( \frac{1}{3} \left( \frac{x}{3} \right)^3 - \frac{1}{3} \left( \frac{3}{x} \right)^3 \right) \] \[ = \frac{1}{x} \left( \frac{1}{3} \cdot \frac{x^3}{27} - \frac{1}{3} \cdot \frac{27}{x^3} \right) \] \[ = \frac{1}{x} \left( \frac{x^3}{81} - \frac{9}{x^3} \right) \] \[ = \frac{x^2}{81} - \frac{9}{x^4} \] Now integrate with respect to \( x \): \[ M = \int_{x=3}^{6} \left( \frac{x^2}{81} - \frac{9}{x^4} \right) dx \] \[ = \frac{1}{81} \int_{3}^{6} x^2 dx - 9 \int_{3}^{6} x^{-4} dx \] Calculate each term: - \( \int x^2 dx = \frac{x^3}{3} \) - \( \int x^{-4} dx = -\frac{1}{3} x^{-3} \) Plug in the limits: \[ = \frac{1}{81} \left[ \frac{x^3}{3} \right]_{3}^{6} - 9 \left[ -\frac{1}{3} x^{-3} \right]_{3}^{6} \] \[ = \frac{1}{81} \left( \frac{216}{3} - \frac{27}{3} \right) + 3 \left( \frac{1}{216} - \frac{1}{27} \right) \] \[ = \frac{1}{81} (72 - 9) + 3 \left( \frac{1 - 8}{216} \right) \] \[ = \frac{63}{81} + 3 \left( -\frac{7}{216} \right) \] \[ = \frac{7}{9} - \frac{7}{72} \] Find a common denominator: \[ \frac{56}{72} - \frac{7}{72} = \frac{49}{72} \] **Final Answer:** \[ \boxed{\frac{49}{72} \text{ grams}} \] **Explanation**: We evaluated the double integral step-by-step, simplified, and combined the results to find the total mass of the plate.