Question Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of \( f(x)=3 x+1 \) an below by the \( x \)-axis over the interval \( [5,6] \) about the line \( y=-4 \). Enter an exact value in terms of \( \pi \).

To find the volume of the solid of revolution formed by revolving the region bounded by \( f(x)=3x+1 \) and the \( x \)-axis over the interval \( [5,6] \) about the line \( y=-4 \), we can use the disk method. The volume \( V \) can be calculated using the formula: \[ V = \pi \int_{a}^{b} \left( [f(x) - (-4)]^2 - [0 - (-4)]^2 \right) dx \] Substituting in our functions and limits: \[ V = \pi \int_{5}^{6} \left( (3x + 1 + 4)^2 - 4^2 \right) dx \] This simplifies to: \[ V = \pi \int_{5}^{6} \left( (3x + 5)^2 - 16 \right) dx \] Calculating the integral involves expanding \( (3x + 5)^2 \) and integrating term by term. After simplifying and evaluating from 5 to 6, we find: \[ \int_{5}^{6} (3x + 5)^2 dx = \left[ \frac{(3x + 5)^3}{9} \right]_{5}^{6} = \left[ \frac{(23)^3}{9} - \frac{(20)^3}{9} \right] \] After evaluating, we plug back into our volume formula and solve to find the exact volume in terms of \( \pi \): \[ V = \frac{89\pi}{3} \] So, the final answer is \( \frac{89\pi}{3} \). The beautiful intersection of mathematics and history can be seen in the development of calculus during the 17th century. Pioneered by great minds like Isaac Newton and Gottfried Wilhelm Leibniz, calculus was formulated as a method to solve problems involving areas, volumes, and rates of change—essentially laying the groundwork for how we compute volumes of solids of revolution like this one! When it comes to solving problems like these, one common mistake is forgetting to adjust for the axis of rotation, which can easily lead to errors in volume calculation. Remember to visualize the graph and consider how far the function is from the line you’re revolving around—this ensures your integrals are set up correctly and you capture the true dimensions of the solid you're creating.