Uestion fine \( R \) as the region bounded above by the graph of \( f(x)=x+5 \), below by the \( x \)-axis, on the left by \( x=1 \), and on t hht by \( x=4 \). What is the volume of the solid obtained by rotating \( R \) about the line \( y=-4 \) ? Enter an exact value in ter

To find the volume of the solid obtained by rotating the region \( R \) about the line \( y = -4 \), we can use the disk method. First, we identify the function and the limits of integration. The area \( R \) is bounded above by the line \( f(x) = x + 5 \), below by the \( x\)-axis (which is \( y = 0 \)), on the left by \( x = 1 \), and on the right by \( x = 4 \). When rotating around \( y = -4 \), we need to calculate the distance from the line of rotation to the function. The top of the region is at \( y = x + 5 \) and the bottom is at \( y = 0 \). Therefore, the radius \( R(x) \) for the washer method is given by: \[ R(x) = (x + 5) - (-4) = x + 5 + 4 = x + 9 \] Thus, the volume \( V \) can be calculated using the following integral: \[ V = \pi \int_{1}^{4} (R(x))^2 \, dx = \pi \int_{1}^{4} (x + 9)^2 \, dx \] Now we can expand the integrand: \[ (x + 9)^2 = x^2 + 18x + 81 \] Now integrate: \[ \int (x^2 + 18x + 81) \, dx = \frac{x^3}{3} + 9x^2 + 81x \] Next, we evaluate this from \( x = 1 \) to \( x = 4 \): At \( x = 4 \): \[ \frac{4^3}{3} + 9(4^2) + 81(4) = \frac{64}{3} + 144 + 324 = \frac{64}{3} + \frac{432}{3} + \frac{972}{3} = \frac{64 + 432 + 972}{3} = \frac{1468}{3} \] At \( x = 1 \): \[ \frac{1^3}{3} + 9(1^2) + 81(1) = \frac{1}{3} + 9 + 81 = \frac{1}{3} + \frac{27}{3} + \frac{243}{3} = \frac{271}{3} \] Now, subtract the two results: \[ \left(\frac{1468}{3} - \frac{271}{3}\right) = \frac{1468 - 271}{3} = \frac{1197}{3} \] So, the volume \( V \) is given by: \[ V = \pi \cdot \frac{1197}{3} = \frac{1197\pi}{3} \] Thus, the volume of the solid obtained by rotating \( R \) about the line \( y = -4 \) is \( \frac{1197\pi}{3} \) cubic units.