Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis. x + y = 7, x = 16 − (y − 3)2

Answer:[tex]v= \frac{2401\pi}{6}[/tex]Step-by-step explanation:I put the equations in a plotter, you can see the rotation region below. Now, as we are rotationg about the x-axis we are going to use the equations and the limit points in terms of y. Now, the volume is [tex]v = 2\pi\int\limits^a_b {p(y)h(y)} \, dy[/tex] where p(y) is the distance from the rotation axis and the differential, a and b are the limit points of the region and h(y) is the height of the region. In this case p(y)= y, a=0, b=7 and h(y) = 16-(y-3)^2-(7-y) (right minus left). So,[tex]v = 2\pi\int\limits^0_7 {y(16-(y-3)^{2}-7+y)} \, dy[/tex][tex]v = 2\pi\int\limits^0_7 {y(9-(y-3)^{2}+y)} \, dy[/tex][tex]v = 2\pi\int\limits^0_7 {y(9-(y^{2}-6y+9)+y)} \, dy[/tex][tex]v = 2\pi\int\limits^0_7 {y(9-y^{2}+6y-9+y)} \, dy[/tex][tex]v = 2\pi\int\limits^0_7 {y(-y^{2}+7y)} \, dy[/tex][tex]v = 2\pi\int\limits^0_7 {-y^{3}+7y^{2}} \, dy[/tex][tex]v= 2\pi (-\frac{y^{4}}{4}+\frac{7y^{3}}{3})^{7}_0[/tex][tex]v= 2\pi (-\frac{7^{4}}{4}+\frac{7*7^{3}}{3})[/tex][tex]v= 2\pi (-\frac{2401}{4}+\frac{2401}{3})[/tex][tex]v= 2\pi (\frac{2401}{12})[/tex][tex]v= \frac{2401\pi}{6}[/tex].