14. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the given lines. \[ y=\sqrt{x}, y=0, x=4 \] a. Rotated around the \( x \)-axis b. Rotated around the \( y \)-axis c. Rotated around the line \( x=4 \) d. Rotated around the line \( x=6 \)

To find the volume of the solid generated by revolving the region bounded by the curves about the specified lines, we can use the method of disks and shells. First, let's identify the area we are working with. The region is bounded by the curves \(y = \sqrt{x}\), \(y = 0\) (the x-axis), and the vertical line \(x = 4\). This forms a region that looks like a "slice" of a curve. ### a. Rotated around the \(x\)-axis When rotated around the \(x\)-axis, we can use the disk method. The volume \(V\) is given by: \[ V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx = \pi \int_{0}^{4} x \, dx = \pi \left[ \frac{x^2}{2} \right]_{0}^{4} = \pi \left( \frac{16}{2} - 0 \right) = 8\pi \] ### b. Rotated around the \(y\)-axis For rotation about the \(y\)-axis, we will use the shell method. The volume \(V\) is given by: \[ V = 2\pi \int_{0}^{4} x \cdot \sqrt{x} \, dx = 2\pi \int_{0}^{4} x^{3/2} \, dx = 2\pi \left[ \frac{x^{5/2}}{5/2} \right]_{0}^{4} = 2\pi \cdot \frac{2}{5} \cdot (4^{5/2}) = 2\pi \cdot \frac{2}{5} \cdot 32 = \frac{128\pi}{5} \] ### c. Rotated around the line \(x=4\) Using the shell method again, the volume generated is calculated by adjusting the radius. The volume \(V\) now is: \[ V = 2\pi \int_{0}^{4} (4-x) \cdot \sqrt{x} \, dx = 2\pi \int_{0}^{4} (4\sqrt{x} - x^{3/2}) \, dx \] Calculating each integral, we find: \[ = 2\pi \left[ \frac{8}{3} x^{3/2} - \frac{2}{5} x^{5/2} \right]_{0}^{4} = 2\pi \left( \frac{8}{3} \cdot 8 - \frac{2}{5} \cdot 32 \right) = 2\pi \left( \frac{64}{3} - \frac{64}{5} \right) \] This gives a final volume of \(V = \frac{256\pi}{15}\). ### d. Rotated around the line \(x=6\) In this case, where we rotate about \(x=6\), we again use the shell method: \[ V = 2\pi \int_{0}^{4} (6-x) \cdot \sqrt{x} \, dx = 2\pi \int_{0}^{4} (6\sqrt{x} - x^{3/2}) \, dx \] Calculating the integrals gives: \[ = 2\pi \left[ 6 \cdot \frac{8}{3} - \frac{2}{5} \cdot 32 \right] = 2\pi \left( 16 - \frac{64}{5} \right) = 2\pi \left( \frac{80}{5} - \frac{64}{5} \right) = \frac{32\pi}{5} \] In summary, the volumes are: - **a.** \(8\pi\) - **b.** \(\frac{128\pi}{5}\) - **c.** \(\frac{256\pi}{15}\) - **d.** \(\frac{32\pi}{5}\)