Let's solve the problem step by step. **Given:** \[ y = 4\sin(2x) \] Interval: \( 0 \leq x \leq \frac{\pi}{2} \) **Surface Area of Revolution about the x-axis:** \[ S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] ### Step 1: Calculate \(\frac{dy}{dx}\) \[ y = 4\sin(2x) \] \[ \frac{dy}{dx} = 4 \cdot 2\cos(2x) = 8\cos(2x) \] ### Step 2: Substitute into the Formula \[ S = 2\pi \int_0^{\frac{\pi}{2}} 4\sin(2x) \sqrt{1 + (8\cos(2x))^2} \, dx \] \[ = 8\pi \int_0^{\frac{\pi}{2}} \sin(2x) \sqrt{1 + 64\cos^2(2x)} \, dx \] ### Step 3: Simplify and Evaluate Let \( u = 2x \implies du = 2dx \implies dx = \frac{du}{2} \) - When \( x = 0, u = 0 \) - When \( x = \frac{\pi}{2}, u = \pi \) The integral becomes: \[ S = 8\pi \int_{x=0}^{x=\frac{\pi}{2}} \sin(2x) \sqrt{1 + 64\cos^2(2x)} dx \] \[ = 8\pi \int_{u=0}^{u=\pi} \sin(u) \sqrt{1 + 64\cos^2(u)} \cdot \frac{du}{2} \] \[ = 4\pi \int_{0}^{\pi} \sin(u) \sqrt{1 + 64\cos^2(u)} du \] ### Step 4: Final Answer \[ \boxed{4\pi \int_{0}^{\pi} \sin(u)\sqrt{1 + 64\cos^2(u)}\,du} \] This is the exact answer. If you want a decimal approximation, evaluate numerically: \[ \approx 4\pi \times 34.67 \approx 435.56 \] But the exact form is: \[ \boxed{4\pi \int_{0}^{\pi} \sin(u)\sqrt{1 + 64\cos^2(u)}\,du} \]