Inswer above is NOT correct. the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. \[ y=1 / x^{3}, \quad y=0, \quad x=4, \quad x=8 \text {; } \] ut the \( y \)-axis. ume \( =\frac{3 \pi^{2}}{32} \) view My Answers score was recorded. score was successfully sent to the LMS. nave attempted this problem 3 times.

To find the volume of the solid obtained by rotating the region bounded by the curves \( y = \frac{1}{x^3} \), \( y = 0 \), \( x = 4 \), and \( x = 8 \) about the \( y \)-axis, we can use the method of cylindrical shells. ### Step 1: Identify the volume formula The volume \( V \) of the solid of revolution using the shell method is given by the formula: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] where \( f(x) \) is the function being rotated, and \( [a, b] \) is the interval of integration. ### Step 2: Set up the integral In this case: - \( f(x) = \frac{1}{x^3} \) - The limits of integration are \( a = 4 \) and \( b = 8 \). Thus, the volume can be expressed as: \[ V = 2\pi \int_{4}^{8} x \cdot \frac{1}{x^3} \, dx \] ### Step 3: Simplify the integrand The integrand simplifies to: \[ x \cdot \frac{1}{x^3} = \frac{1}{x^2} \] So the integral becomes: \[ V = 2\pi \int_{4}^{8} \frac{1}{x^2} \, dx \] ### Step 4: Calculate the integral Now we can calculate the integral: \[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} \] Thus, we evaluate: \[ V = 2\pi \left[ -\frac{1}{x} \right]_{4}^{8} \] Calculating the definite integral: \[ V = 2\pi \left( -\frac{1}{8} + \frac{1}{4} \right) \] \[ = 2\pi \left( -\frac{1}{8} + \frac{2}{8} \right) \] \[ = 2\pi \left( \frac{1}{8} \right) \] \[ = \frac{2\pi}{8} = \frac{\pi}{4} \] ### Step 5: Final volume calculation Now, we multiply by 2: \[ V = 2\pi \cdot \frac{1}{8} = \frac{\pi}{4} \] ### Conclusion The volume of the solid obtained by rotating the region bounded by the given curves about the \( y \)-axis is: \[ V = \frac{\pi}{4} \] It seems there was a misunderstanding in the initial volume calculation. The correct volume is \( \frac{\pi}{4} \), not \( \frac{3\pi^2}{32} \).