Let's work step-by-step to solve the problem: ## **Problem Restatement** Use integration to find the lateral surface area of a right circular cone of height 4 and radius 3. **First:** Indicate in words how the surface is generated. **Second:** Set up the integral. **Third:** Evaluate the integral. --- ## **Step 1: Describe How the Surface is Generated** A right circular cone can be generated by rotating a straight line segment (from the apex to the base) about the vertical axis (the height of the cone). The lateral surface is the curved surface that wraps around the side, excluding the base. --- ## **Step 2: Set Up the Integral** Let’s model the cone with its vertex at the origin \((0, 0)\), and its base at \(x = 4\) (height = 4), with the radius at \(x = 4\) being 3. The straight line from the apex to the edge of the base has endpoints at \((0, 0)\) and \((4, 3)\). **Equation of the line:** \[ y = mx \] where \(m = \frac{3}{4}\). So, \(y = \frac{3}{4}x\). If we revolve this line about the x-axis from \(x = 0\) to \(x = 4\), it sweeps out the cone. ### **Surface Area of Revolution Formula** The surface area \(A\) generated by revolving a curve \(y = f(x)\) from \(x = a\) to \(x = b\) about the x-axis is: \[ A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] **Here:** - \(y = \frac{3}{4}x\) - \(a = 0\), \(b = 4\) - \(\frac{dy}{dx} = \frac{3}{4}\) Plug these into the formula: --- ## **Step 3: Setup and Evaluate the Integral** Plug in values: \[ A = 2\pi \int_{0}^{4} \left(\frac{3}{4}x\right) \sqrt{1 + \left(\frac{3}{4}\right)^2} \, dx \] Calculate the square root: \[ \left(\frac{3}{4}\right)^2 = \frac{9}{16} \] So, \[ \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \] Now plug in: \[ A = 2\pi \int_{0}^{4} \left(\frac{3}{4}x\right) \cdot \frac{5}{4} dx \] \[ A = 2\pi \cdot \frac{3}{4} \cdot \frac{5}{4} \int_{0}^{4} x \, dx \] \[ = 2\pi \cdot \frac{15}{16} \int_{0}^{4} x \, dx \] Integrate \(x\) from 0 to 4: \[ \int_{0}^{4} x \, dx = \left. \frac{x^2}{2} \right|_0^4 = \frac{16}{2} = 8 \] So, \[ A = 2\pi \cdot \frac{15}{16} \cdot 8 \] \[ = 2\pi \cdot \frac{120}{16} \] \[ = 2\pi \cdot 7.5 = 15\pi \] --- ## **Final Answer** \[ \boxed{15\pi} \] **The lateral surface area of the cone is \(15\pi\) square units.** --- ### **Summary Table** | Step | Explanation | |------|-------------| | 1 | The surface is generated by rotating the slant height (a line from the apex to the base edge) about the axis. | | 2 | Set up the integral for the surface area of revolution using \(y = \frac{3}{4}x\) from \(x = 0\) to \(x = 4\). | | 3 | Solve: \(A = 2\pi \int_{0}^{4} \left(\frac{3}{4}x\right)\left(\frac{5}{4}\right)dx = 15\pi\) | Let me know if you need a diagram or further clarification!