Volumes of Solids of Revolution. We revolve this curve about the x-axis and create a solid of revolution.

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1 Volumes of Solids of Revolution Consider the function ( ) from a = to b = We revolve this curve about the x-axis and create a solid of revolution We want to find the volume of this solid. We begin by taking n cross-sectional slices of width.

2 Page The shape of one of these slices can be approximated by a disk. We will find the formula for the volume of one of these disks and then use Riemann sums and limits to produce an integral that will give us the exact volume of our solid of revolution. Laying a disk on its side, compute its volume as a cylinder. V = area of the base * height Note that for this disk, r is the distance between the x-axis and the curve. That is, for this disk, ( ) for some x in that slice. Likewise,. So the volume of this disk is ( ( )). Then when we approximate the volume of the solid of revolution using disks, the volume of the ith disk is ( ( )). The sum of the volumes of n disks is a Riemann Sum: ( ( )). Taking the limit as n goes to infinity, we get the integral ( ( )). Thus the integral for the volume of the solid of revolution generated by revolving the curve ( ) from x = a to x = b about the x-axis is ( ( )).

3 Page So, the volume of the solid of revolution that is generated by revolving the curve ( ) from a = to b = 9 about the x-axis is ( ). Ah, but what if we revolve ( ) about the y-axis? We will get to that on p. 5. But first, what if we revolve the region between two curves about the x-axis? For example, consider the region between ( ), y= and the y-axis When we revolve this about the x-axis, we will get something like a vase.

4 Page If we were to take cross-sectional slices, we would get something shaped like washers: disks with a smaller disk cut out of the center: The volume of a washer is the volume of the larger disk minus the volume of the inner disk that has been removed. Larger disk: V = area of base * height = where r =. Inner disk: V = area of base * height = where r =. So, the volume of this solid of revolution, using the method of washers is ( ) So, if ( ) ( ) on the interval [ ] then the volume of the solid of revolution generated by revolving the region enclosed by ( ) ( ) and is ( ( )) ( ( )) and, using properties of integrals, this is the same as (( ( )) ( ( )) )

5 Page 5 We now return to the question of what if we revolve ( ) about the y-axis: There are two possible ways to approach this. If ( ) is a one-to-one function on [a, b] and we can write x as a function of y, we can still use the method of disks, but our cross-sectional slicing is horizontal. For each slice, the approximating disk s radius is the distance from the y-axis to the curve: ( ) and the thickness of the disk is dy. The variable y is going from ( ) to ( ), so the volume is ( ( )) In general, using disks when revolving ( ) about the y-axis: ( ( )) where c and d are the lower and upper limits on y. In the problem above, revolving our function from a = to b = 9 about the y-axis, we would solve for x, giving us, then change our limits of integration from and b and and d: Lower limit: ( ) Upper limit: ( ). ( ). But if ( ) is not a one-to-one function on [a, b] or if we cannot write x as a function of y, or we are unable to evaluate the resulting integral, we cannot use the method of disks, so another method will have to be used.

6 Page 6 Using the solid from the previous example: This time, we take vertical cylindrical slices, sort of like taking the core out of a pineapple, except we take thin cylindrical slices We approximate these with perfectly cylindrical shells. When we examine one of these cylindrical shells, we see that we can cut it vertically and lay it flat as a rectangular slab. Volume of the cylindrical shell = volume of slab: The width of the rectangle is the circumference of the base of the cylinder:. Looking at the cylindrical slice while it is still inside the solid, we see that r = x. So, Next,., so. Finally,. So the volume of the cylindrical shell is ( ).

7 Page Take the Riemann sum of the volumes of n cylindrical shells: ( ) Then take the limit as n goes to infinity, and we have the integral ( ) = ( ) = So, using the method of cylindrical shells, ( )