Dr. Z’s Math152 Handout #6.3 [Volumes by Cylindrical Shells]By Doron ZeilbergerProblem Type 6.3a: Use the method of cylindrical shells to find the volume generated by rotatingthe region bounded by the given curves about the y-axis.y = f(x) , y = 0 , x = a, x = bExample Problem 6.3a: Use the method of cylindrical shells to find the volume generated byrotating the region bounded by the given curves about the y-axis.y = 1/x2, y = 0 , x = 1 , x = 2Steps Example1. Sketc the region and make sure thatindeed x = a and x = b (that are givenby the problem) bound the region, andthat the x-axis (alias y = 0) is below thecurve y = f(x). a and b are your limits ofintegration.1. Sketching the region (do it!) showsthat indeed x = 1 and x = 2 are the limitsof integration.2. Set-up the integral with the folowingformula.V olume = 2πZbaxf(x)dx2.V olume = 2πZ21x1x2dx3. Evaluate the integral. 3.V olume = 2πZ21x1x2dx = 2πZ211xdx =2π(ln x)|21= 2π(ln 2−ln 1) = 2π(ln 2−0) = 2π ln 2 .Ans.: 2π ln 21Problem Type 6.3b: Use the method of cylindrical shells to find the volume generated by rotatingthe region bounded by the given curves about the y-axis.y = f(x) , y = g(x) .Example Problem 6.3b: Use the method of cylindrical shells to find the volume generated byrotating the region bounded by the given curves about the y-axis.y = 3 + 2x − x2, x + y = 3 .Steps Example1. Sketch the two curves and find theregion bounded by them. Unlike 6.3a,where the limits of integration a and bare given to you, here you must find themby solving for x the equation f(x) = g(x).You should expect to get two roots. Theseare your limits of integration a and b. Lookat the sketch and decide who is on TOPand who is at the BOTTOM.1. In this case the second curve (thathappens to be a line) is given implictly,and you must first convert it to explicitform y = Expression(x). In this case,x + y = 3 becomes y = 3 − x. Solving3 + 2x − x2= 3 − x yields x2− 3x = 0,which is x(x − 3) = 0 giving the two rootsx = 0 and x = 3. These are your limitsof integration. From the sketch (do it!)T OP = 3 + 2x − x2and BOT T OM =3 − x.2. Set-up the integral with the followingformula.V olume = 2πZbax(T OP −BOT T OM )dx2.V olume = 2πZ30x(3 + 2x − x2) − (3 − x)dx23. Evaluate the integral. 3.V olume = 2πZ30x(3 + 2x − x2) − (3 − x)dx= 2πZ30x(3x − x2)dx= 2πZ30(3x2− x3)dx = 2π(x3− x4/4)|30= 2π((33−34/4)−0) = 2π(27/4) = 27π/2