we make slices perpendicular to the x-axis. If the slices are thin enough, they resemble x cylinders or discs. The formula for the x

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1 Math Learning Centre Solids of Revolution When we rotate a curve around a defined ais, the -D shape created is called a solid of revolution. In the same wa that we can find the area under a curve calculating the integral of the curve s equation, we can also use integration to calculate the volume of a space swept out (created) rotating a curve around an ais of rotation. The principle is the same: we chop the volume up into man ver thin slices and add up the volumes of the slices to approimate the total, and then use the integral to see what happens with an infinite numer of slices. Since the solid has three dimensions, we have two options: we can slice perpendicular or parallel to the ais of rotation. If the slices are perpendicular to the ais we make discs (or washers if the solid has a hole or depression in it), and if the slices are parallel to the ais we make clindrical shells. Since the ais of rotation might e either horiontal or vertical, and the relation that defines the curve might e more easil epressed in terms of or, we will decide whether taking perpendicular vs. parallel slices it the est approach for prolem solving. PERPENDICULAR SLICING DISCS AND WASHERS We have a continuous curve, ounded in Quadrant I, and it s epressed in terms of = ƒ(). We are asked to rotate this curve around the -ais to form a solid, and determine the volume of this solid. The graph of the curve and the solid are shown elow: The original curve is in old in the -D view. The solid is a ullet shape with a perfectl circular ack. To find the volume of the solid, = ƒ() we make slices perpendicular to the -ais. If the slices are thin enough, the resemle clinders or discs. The formula for the volume of a clinder is π (radius)² (height). (a, ) The height for each disc will e the infinitel small width of a slice, which is represented d. The radius is the distance from the a point on the ais of rotation to the point on the curve that has the same -coordinate, which is ƒ(). We ll e making slices along the ais of rotation from the start to the finish of the solid. In this case, it starts at origin ( = ) and finishes where the curve intersects the ais ( = a). This means the total volume of the slices is π [ƒ()]² d for all values of from to a. When we cut an infinite numer of slices, the sum ecomes a Riemann sum of the discs volumes, which is epressed as an integral: V = π [ƒ()]² = π [ƒ()]². = ƒ() Vancouver Communit College Learning Centre. Authored Emil Darren Simpson Rig Student review onl. Ma not e reproduced for classes. (a, )

2 If the ais of rotation isn t the -ais, ut some other line = k, then we have to adjust the value ƒ() to account for the change in radius. We do this using the same method as in algeraic transformations: we re altering the value, so we sutract k from ƒ(): V = π If ais is elow: [ƒ() k]² d If ais is aove: = ƒ() In oth cases, the radius of the curve at an point is ƒ() k. Note: [ƒ() k]² is not the same thing as ƒ()² k². Now consider the possiilit that the curve is not ounded the aes and the ais of rotation is not the - or -ais. If we have the same curve ƒ() ounded the lines = m and =, and rotated around the line = k, where k < m: This shape has a clindrical hole in its centre. We can calculate the volume of this new shape finding the volume of the shape without the hole, and then sutracting the volume of the hole. = k (a, k) (, k) = = ƒ() (c, k) = ƒ() (a, k) = m = k = k In general, if the function that defines the inside surface of such a shape is g(), then: V = V outside V inside = π [ƒ() k]² d π [g() k]² d = π { [ƒ() k]² d [g() k]² d } In this particular eample, g() = m, so the inside radius is constant, m k. If the ais of rotation has een moved to the other side of the curve as the - or -ais (as appropriate), then the adjustment in these formulas is reversed: [k ƒ()]. The slices in the Riemann sum associated with this integral are circular with a hole in them, so the are sometimes called washers, after the metal part ou find on a olt. All of these eamples might just as easil have een descried as a relation = ƒ() and the resulting curves rotated around vertical aes. In this case the s and s in the formulas and integrals would e switched. Everthing else remains the same. Vancouver Communit College Learning Centre. Student review onl. Ma not e reproduced for classes.

3 PARALLEL SLICING CYLINDRICAL SHELLS Consider a curve, = ƒ(), on an interval [, a] whose inverse would not e a function. Let s sa for simplicit s sake that a is a ero of ƒ(). The curve is rotated around the - ais to create a solid of revolution as shown elow: We could make slices perpendicular to the ais of rotation since our ais of rotation is vertical, we d make horiontal slices and integrate with respect to ut the inverse relation ƒ () is inconvenient to work with ecause it introduces an interior surface for part of the solid. It s usuall difficult to separate the interior and eterior sections of the curve to write the integral. It would e etter to integrate with respect to (using slices parallel to the ais of rotation) so we avoid this prolem. Our slices would then e rotated just like the curve was, descriing paper-thin clinders, or clindrical shells. Just like a hollow clinder made out of a rectangle of paper, we can approimate the volume of the shell finding the volume of the piece of paper (a rectangular solid) that we would roll up to get the shell. The volume of the shell, then is (length) (width) (height). The width of the shell is d. The height of an given shell is ƒ(). The length will e the circumference of the shell, which is π times the radius. The radius is the value at the shell s position. The total volume of the solid is the sum of the volumes of all the shells with radii from to a. With an infinite numer of slices, we get an integral. The formula ecomes: V = π ƒ() d = π ƒ() d If the ais of rotation isn t the -ais, ut some other aritrar vertical line, = h, then the height of each shell doesn t change, ut the radius does: V = π ( h) ƒ() d If the ais has een shifted to the other side (or outside) of the curve, the radius should e epressed as (h ) instead. For a relation = g() that s rotated around a horiontal ais, the same strateg applies, and all the s and s switch places. There is no analogue to washers for the clindrical shells method of finding the volume. We tend to use clindrical shells to integrate parallel to the ais ecause the curve isn t a function in the other direction. If there are holes in the direction of integration as well, then the relation isn t a function in an direction! So which strateg do ou use for an given question? There s no firm rule. Sometimes it s much easier to integrate ƒ () than it is [ƒ()]², or vice versa. Sometimes, as in the shape aove for clindrical cells, there s a partial depression created a curve that overlaps itself in one direction (i.e., it fails the Vertical or Horiontal Line Test, as appropriate); in that case, use the other direction. Use our etter judgment, and if slicing in one direction doesn t work, tr the other one. Vancouver Communit College Learning Centre. Student review onl. Ma not e reproduced for classes. = ƒ() (a, )

4 Eample : Find the volume of the solid of revolution ounded the curves = and = ³, and rotated around the ais =. Solution: First we sketch the graph and find the points of intersection. The graph is at right, and we get the intersection points (, ) and (, ) inspection. Which sstem should we use? Since these are oth simple power functions, and the re oth alread written in terms of, we should e ale to use clindrical shells. V = π ( h) ƒ() d The ais of rotation is at =, so h =. The height of the shaded shape at an value of on [, ] is the difference etween the -value for the curve that defines the top of the shape, =, and the - value for the curve that defines the ottom of the shape, = ³. We work these into the integral equation and evaluate: V = π ( h) ƒ() d = π ( + ) ( ³) d = π / 4 + / d = π / + / 4 = π + ( + ) = If the question specificall asked that we use washers, we must change each function into its inverse. The curve that defines the outside of the shape (farther from the ais of rotation) ecomes = ³ and the curve that defines the inside of the shape (closer to the ais of rotation) ecomes = ². V = π [ƒ() h]² [g() h]² d = π ( ³ + )² (² + )² d = π ³ ² + ³ + 4 ² d = π / + / 4 d = π / + 4/ = π + ( + ) = 7π 7π The answers agree, which is what we would epect. 4 4 Vancouver Communit College Learning Centre. Student review onl. Ma not e reproduced for classes. 4

5 EXERCISES A. Find the volume of the solid of revolution formed rotating the region ounded the listed curves aout the given ais. Use the method of discs or washers. ) + =, =, = ; aout the -ais ) = ln, =, =, = ; aout the -ais ) = sec where ( π, π ), = ; aout the -ais 4) =, =, = ; aout = ) = ², = ² + ; aout = B. Find the volume of the solid of revolution formed rotating the region ounded the listed curves aout the given ais. Use the method of clindrical shells. ) = ³, = + 4, = ; aout the -ais ) =, = 4, = ; aout the -ais ) = 4 ², = ; aout = 4) = + ², =, =, = ; aout the -ais ² ) = on [, ), =, = ; aout = C. Find the volume of the solid of revolution formed rotating the region ounded the listed curves aout the given ais. Use whichever method is more appropriate. ) = 4² + 8, = ; aout the -ais ) ² + ( 4)² = 9, =, = 4; aout the -ais ) = sin ², [, ), = ; aout the -ais D. ) Find the volume of the solid of revolution ounded the curves = ³ ² + and = on the interval [, ], rotated aout the -ais, using the method of clindrical shells. ) Do the same question, ut on the interval [, ]. Compare this answer to D. Is there anthing strange? ) Find the correct volume for the solid on the interval [, ] splitting the integral in an appropriate place. 4) Find the volume of the solid of revolution ounded the curves = ( + )² and = ³ + rotated aout =. SOLUTIONS A: () π () π (e 6 ) () 8π² π (4) π( 8 ln ) () 44π B: () 8π () 6π () 8π (4) π (4 ) () π(ln ) C: () discs: 6π () either method: 8π () clindrical shells: π D: () 7π () 8π ; The volume has decreased. This happened ecause the curves intersected at =. The volume included in D was all negative. () 7π (4) 69π Vancouver Communit College Learning Centre. Student review onl. Ma not e reproduced for classes.