Let's carefully analyze and solve the problem step-by-step. --- ### **Problem Restatement:** Find the **center of mass** \((\bar{x}, \bar{y})\) of a thin plate with density \(\delta = 3\), bounded by: - the lines \(y = x\), - the line \(x=0\), - and the parabola \(y = 30 - x^2\), in the **first quadrant**. --- ### **Step 1: Understand the region** - The region is bounded on the left by \(x=0\), - below by \(y = x\), - above by \(y = 30 - x^2\), - and in the first quadrant, so \(x \ge 0\), \(y \ge 0\). Let's find the intersection points of the curves to determine the limits: Solve \(x = 30 - x^2\): \[ x = 30 - x^2 \Rightarrow x^2 + x - 30 = 0. \] Using quadratic formula: \[ x = \frac{-1 \pm \sqrt{1 + 120}}{2} = \frac{-1 \pm \sqrt{121}}{2} = \frac{-1 \pm 11}{2}. \] Possible solutions: - \(x = \frac{-1 + 11}{2} = \frac{10}{2} = 5\), - \(x = \frac{-1 - 11}{2} = -6\) (discarded since outside first quadrant). So, the intersection point in the first quadrant is at \(x=5\). At \(x=5\): \[ y = x = 5, \] and also: \[ y = 30 - 5^2 = 30 - 25 = 5, \] which matches, confirming the intersection point is \((5,5)\). --- ### **Step 2: Describe the region** The region \(R\) can be described as: \[ 0 \le x \le 5, \] and for each fixed \(x\), \[ \text{bottom boundary}: y = x, \] \[ \text{top boundary}: y = 30 - x^2. \] Since \(y = x \le 30 - x^2\) for \(x\) in \([0, 5]\), the region's vertical cross-section runs from \(y=x\) up to \(y=30 - x^2\). --- ### **Step 3: Set up integrals for mass and moments** The density \(\delta = 3\) is constant. - **Mass \(M\):** \[ M = \iint_{R} \delta \, dA = 3 \times \text{area of } R. \] - **Coordinates of the center of mass:** \[ \bar{x} = \frac{1}{M} \iint_{R} x \delta \, dA, \] \[ \bar{y} = \frac{1}{M} \iint_{R} y \delta \, dA. \] Since \(\delta\) is constant, it cancels out in the ratio, but for completeness, we will include it. --- ### **Step 4: Set up the double integrals** Expressed in terms of \(x\) and \(y\): \[ M = 3 \int_{x=0}^{5} \int_{y=x}^{30 - x^2} dy\, dx, \] \[ \Rightarrow M = 3 \int_{0}^{5} [ (30 - x^2) - x ] dx, \] because the integral over \(y\) is just the length of the vertical segment for each \(x\). Similarly, \[ \iint_{R} x \delta \, dA = 3 \int_{0}^{5} \int_{y=x}^{30 - x^2} x \, dy\, dx, \] \[ \iint_{R} y \delta \, dA = 3 \int_{0}^{5} \int_{y=x}^{30 - x^2} y \, dy\, dx. \] --- ### **Step 5: Compute the mass \(M\)** Calculate: \[ M = 3 \int_{0}^{5} [ (30 - x^2) - x ] dx = 3 \int_0^5 (30 - x^2 - x) dx. \] Integrate term-by-term: \[ \int_0^5 30\, dx = 30x \big|_0^5 = 150, \] \[ \int_0^5 x\, dx = \frac{x^2}{2} \big|_0^5 = \frac{25}{2} = 12.5, \] \[ \int_0^5 x^2\, dx = \frac{x^3}{3} \big|_0^5 = \frac{125}{3} \approx 41.6667. \] Putting it together: \[ M = 3 \left[ 150 - \frac{125}{3} - 12.5 \right] = 3 \left[ 150 - 41.6667 - 12.5 \right]. \] Compute inside brackets: \[ 150 - 41.6667 = 108.3333, \] \[ 108.3333 - 12.5 = 95.8333. \] Finally, \[ M = 3 \times 95.8333 = 287.5. \] --- ### **Step 6: Compute \(\iint x \delta \, dA\)** \[ \iint_{R} x \delta \, dA = 3 \int_0^5 x \left( (30 - x^2) - x \right) dx, \] \[ = 3 \int_0^5 x (30 - x^2 - x) dx, \] \[ = 3 \int_0^5 (30x - x^3 - x^2) dx. \] Integrate term-by-term: \[ \int_0^5 30x dx = 15 x^2 \big|_0^5 = 15 \times 25 = 375, \] \[ \int_0^5 x^3 dx = \frac{x^4}{4} \big|_0^5 = \frac{625}{4} = 156.25, \] \[ \int_0^5 x^2 dx = \frac{x^3}{3} \big|_0^5 = \frac{125}{3} \approx 41.6667. \] Putting it all together: \[ \iint_{R} x \delta dA = 3 (375 - 156.25 - 41.6667) = 3 (177.0833) = 531.25. \] Now, the \(x\)-coordinate of the center of mass: \[ \boxed{ \bar{x} = \frac{1}{M} \times \iint x \delta dA = \frac{531.25}{287.5} = \frac{531.25}{287.5}. } \] Calculate: \[ \bar{x} \approx 1.8478. \] --- ### **Step 7: Compute \(\iint y \delta \, dA\)** \[ \iint_{R} y \delta dA = 3 \int_0^5 \int_{y=x}^{30 - x^2} y\, dy\, dx. \] Inner integral: \[ \int_{y=x}^{30 - x^2} y\, dy = \frac{1}{2} y^2 \big|_{y=x}^{30 - x^2} = \frac{1}{2} [ (30 - x^2)^2 - x^2 ]. \] Compute: \[ (30 - x^2)^2 = 900 - 60 x^2 + x^4, \] so: \[ \int_{y=x}^{30 - x^2} y\, dy = \frac{1}{2} [900 - 60 x^2 + x^4 - x^2] = \frac{1}{2} (900 - 61 x^2 + x^4). \] Multiply by 3: \[ \iint_{R} y \delta dA = 3 \times \frac{1}{2} \int_0^5 (900 - 61 x^2 + x^4) dx = \frac{3}{2} \int_0^5 (900 - 61 x^2 + x^4) dx. \] Compute the integral term-by-term: \[ \int_0^5 900\, dx = 900 \times 5 = 4500, \] \[ \int_0^5 61 x^2\, dx = 61 \times \frac{x^3}{3} \big|_0^5 = 61 \times \frac{125}{3