Let's solve the problem step by step: --- ### **Step 1: Parameterize the Curve** The rod lies along the first-quadrant portion of the circle \( x^2 + y^2 = 36 \). - The circle has radius 6 and center at (0,0). - In the **first quadrant**, \( x \geq 0, y \geq 0 \). - Parameterize using \( x = 6\cos\theta, \ y = 6\sin\theta \), where \( 0 \leq \theta \leq \frac{\pi}{2} \). **Explanation:** Parameterizing the curve helps express everything in terms of a single variable (\(\theta\)), which is useful for integration. --- ### **Step 2: Set Up the Center of Mass Formulas** The density is \( \rho = 6\sin\theta \). The center of mass coordinates for a curve are: \[ \bar{x} = \frac{1}{M} \int x \rho \, ds \] \[ \bar{y} = \frac{1}{M} \int y \rho \, ds \] where \( ds \) is the element of arc length: \( ds = \sqrt{(dx/d\theta)^2 + (dy/d\theta)^2} d\theta \). Calculate \( ds \): \[ dx/d\theta = -6\sin\theta, \quad dy/d\theta = 6\cos\theta \] \[ ds = \sqrt{(-6\sin\theta)^2 + (6\cos\theta)^2} d\theta = \sqrt{36\sin^2\theta + 36\cos^2\theta} d\theta = 6 d\theta \] **Explanation:** We substitute everything in terms of \(\theta\) and simplify the arc length element. --- ### **Step 3: Compute Mass and Center of Mass Coordinates** **Total mass:** \[ M = \int_0^{\frac{\pi}{2}} \rho \, ds = \int_0^{\frac{\pi}{2}} 6\sin\theta \cdot 6 d\theta = 36 \int_0^{\frac{\pi}{2}} \sin\theta d\theta = 36 [-\cos\theta]_0^{\frac{\pi}{2}} = 36[0 - (-1)] = 36 \] **x-coordinate:** \[ \bar{x} = \frac{1}{36} \int_0^{\frac{\pi}{2}} (6\cos\theta) \cdot 6\sin\theta \cdot 6 d\theta = \frac{1}{36} \int_0^{\frac{\pi}{2}} 216\cos\theta\sin\theta d\theta = \frac{216}{36} \int_0^{\frac{\pi}{2}} \cos\theta\sin\theta d\theta \] \[ = 6 \int_0^{\frac{\pi}{2}} \cos\theta\sin\theta d\theta = 6 \int_0^{\frac{\pi}{2}} \frac{1}{2} \sin(2\theta) d\theta = 3 \int_0^{\frac{\pi}{2}} \sin(2\theta) d\theta = 3 \left[ -\frac{1}{2} \cos(2\theta) \right]_0^{\frac{\pi}{2}} \] \[ = 3 \left( -\frac{1}{2} [\cos(\pi) - \cos(0)] \right) = 3 \left( -\frac{1}{2} [-1 - 1] \right) = 3 \left( -\frac{1}{2} \times -2 \right) = 3 \times 1 = 3 \] **y-coordinate:** \[ \bar{y} = \frac{1}{36} \int_0^{\frac{\pi}{2}} (6\sin\theta) \cdot 6\sin\theta \cdot 6 d\theta = \frac{1}{36} \int_0^{\frac{\pi}{2}} 216 \sin^2\theta d\theta = 6 \int_0^{\frac{\pi}{2}} \sin^2\theta d\theta \] Recall: \[ \int_0^{\frac{\pi}{2}} \sin^2\theta d\theta = \frac{1}{2} \int_0^{\frac{\pi}{2}} (1 - \cos(2\theta)) d\theta = \frac{1}{2} \left[ \theta - \frac{1}{2} \sin(2\theta) \right]_0^{\frac{\pi}{2}} = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4} \] So, \[ \bar{y} = 6 \cdot \frac{\pi}{4} = \frac{3\pi}{2} \] **Explanation:** We integrated to find the mass and both coordinates of the center of mass. --- ## **Final Answer:** \[ \boxed{\bar{x} = 3, \quad \bar{y} = \frac{3\pi}{2}} \] So the correct option is **D**.

The concept used involves parameterizing the curved rod along a circle in the first quadrant using an angle \(\theta\), which simplifies the calculation of mass and moments. By expressing the coordinates, density, and arc length in terms of \(\theta\), we set up integrals for the total mass and the coordinates of the center of mass. Integrating these expressions over the appropriate limits allows us to find the weighted average positions, giving the center of mass of the rod.