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Integrals Go 3D
Shell Volume Integrals

 

 There is a very interesting relationship 

 between the equation for the surface area of a tube ...
 

 A = 2 p r h

 
 and the equation for the volume of a cylinder ...
 

 V = p r2h

 
 Let's rearrange the order of the terms
 to make the relation more obvious.
 

 A = p h 2r

 V = p h r2

 
 See it?
 
 Sure you do!
 It's ...
 

 

 
 in general ...
 

 
 Wow! The equation for the volume of a cylinder
 (of radius "r") is the integral of the area of a tube as "r".
 the radius, goes from zero to some "r" value.
 
 We can use this idea to find the volume of
 round 3 dimensional objects.
 What we need to know is ...
 
 1) The distance from the center of the object
 to the edge (the radius).
 
 2) The height of the object.
 
 A way to think of this is a cylinder,
 made up of very thin tubes (sometimes called shells)
 sitting inside each other like a weird onion.
 

 
 If we were to look at this from the side
 and set the thing on a coordinate axis
 we get something like ...
 

 
 The radius, "r", of each of these tubes is a certain distance
 in the X direction
 so now instead of using "r" in the equation,
 now we could also use X.
 The height (h) of the cylinder is some Y distance,
 so instead of using h in the equation,
 we could also use Y.
 Using X's and Y's the equation becomes ...
 

 
 Here Y is just some constant height,
 so we can write ...
 

 
 This is the volume of the entire cylinder.
 
 What happens if we have a cylinder
 that puts on antigravity shoes and floats up off the X axis?
 

 
 Hey! No problem! 
 We just find out where the top of the cylinder is (call it Y2)
 and where the bottom of the cylinder is (call it Y1
 and instead of Y in the equation we have (Y2 - Y1).
 

 
 Now we have ...
 

 
 (Y1 and Y2 are still just constants)
 
 OK, let's get really wild!
 Suppose the top and bottom of this round thing
 aren't straight lines like Y = 3 or something?
 What if they are functions?
 What if the top is Y = X 2 + 4
 and the bottom is Y = 2X 2
 

 
 This forms a round thing that looks sort of like a shallow bowl
 (but it wouldn't sit on a table too well)
 
 To find the volume of this shape, substitute X 2 + 4 for Y2.
 It is the equation of the upper surface boundary line.
 Substitute 2X 2 for Y1.
 It is the equation of the lower surface boundary line.
 
 We use the place where the two equation lines cross on the right
 as the upper limit of integration.
 Let's find it.
 

 Y = X2 + 4     Y = 2X2

 2X2 = X2 + 4

 X2 = 4

 X = 2

 
 So the equation is ...
 

 
 THAT'S THE EQUATION FOR THE VOLUME OF MATERIAL
 OUR FUNNY CUP IS MADE OUT OF!
 
 Don't get confused and think that it's the equation
 for the volume of stuff you can put IN the cup,
 that's some other equation.
 So let's figure out the volume of the material
 that the cup is made of ...
 

 
 We move the constants (p and 2) left,
 and multiply everything else out.
 

 
 Integrate and solve ...
 

 

 
 In general, if we have any shape like this ...
 

 
 The equation for the volume of material is ...
 

 
 Since 2p is a constant, we can make this ...
 

 

 

 
 We can also make a shape
 out of two lines that don't touch the Y axis.
 Suppose we have ...
 

 

 
 NO BIG DEAL!
 
 Just use "b" as your upper integration limit,
 and "a" (not zero) as the lower limit.
 

 
 This gives us a shape that looks something like
 a donut or an onion ring.
 

 
 One thing to mention, is that the view shown 
 is actually slightly looking down on the shape.
 If we looked at it straight from the side,
 the view would be pretty boring.
 

 
 What if we wanted to make an object like a donut
 that is NOT centered around the Y axis?
 What if we wanted it centered around the line X = -2?
 

 
 It's about the same shape, but much bigger around.
 The radius has been increased.
 
 HOW MUCH?
 
 Well we moved the center of the donut 2 units,
 so I'd say the radius is 2 units bigger.
 
 Before, when the donut was centered on the Y axis
 we used the value of X for the radius distance.
 Now it looks like we use X + 2!
 
 Big Time Example:
 
 Say we want to find the volume of some object
 that is bounded on the top by the line ...
 

 F(X) = - (X - 2)2 + 2

 
 and on the bottom by the line ...
 

 G(X) = (X - 2)2

 
 And we want to rotate it around the line X = -3.
 

 

 
 First we need to find the places that the two lines cross,
 also known as the upper and lower limits of integration (X = a and X = b) ...
 
 (X - 2)2 =  -(X - 2)2 + 2
 X2 - 4X + 4 =  -X2 + 4X - 2
 2X2 - 8X + 6 =  0
 X2 - 4X + 3 =  0
 (X - 3)(X - 1) =  0
 X = 3   or   X = 1
 (b = 3,  a = 1)
 
 We can see from the picture that the radius is X + 3,
 so we use X + 3 for "r" instead of just X.
 

 
 First a bunch of algebra ...
 

 

 
 Now the integration!
 

 

 
 So the volume of this "donut" is 83.8 cubic inches.
 There is a "quick and dirty" way to estimate these puppies ...
 When you are working on a problem that takes this many steps to solve,
 it is easy to make a simple math error and get a very wrong answer.
 It would be great if we could find a way to estimate the answer.
 Then when we finish the big calculation,
 we can see if the answer we got was reasonable.
 
 OK, How do we do it?
 
 Say we have an object with a cross section of:
 

 
 Who knows exactly what the area of this thing is,
 but the square that surrounds it has an area of 4.
 So let's guess that the donut has an area of about 3.
 

 
 The radius of the center of the object is 5,
 so if we cut this donut and straightened it out
 into a cylinder it would have a height equal
 to the distance around the object.
 (circumference = 2pr, and r = 5)
 The volume of this cylinder is equal to 
 the base times the height.
 

 V =  

bh

 V =  

3 10 p

V =  

30 p

V =  

94.2
 
 Our guess was 94.2
 and the actual area was 83.8.
 

   copyright 2005 Bruce Kirkpatrick

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