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We know how to integrate our favorite flat foods to find their areas. Our next step is to learn how to integrate our favorite 3-D foods to find their volumes. In terms of the methods we'll learn, there are two types: solids with known cross-sections and solids of revolution. With these methods, we can find the volume of any 3-D food we want. 4-D foods like the hypergeometric cotton candy will have to wait.
A solid with known cross-sections is like a loaf of cinnamon raisin bread. There may be a lot of things going on within each delicious slice. We're going to slice it that way, nonetheless. The base of the solid is some 2-D region. For this loaf of bread, we can think of the base as a circle in the xy-plane:
When we slice the loaf perpendicular to the x-axis, we get slices that look like semi-circles. The size of the semi-circle depends where along the base the slice was cut. Slices in the middle of the loaf will be bigger and have more cinnamon than slices at the ends. We only care about the area of the slice.
A solid of revolution is a little different from cinnamon raisin bread. Let's imagine a very, very thin piece of bundt cake. That slice, flavored like key lime in this case, is just a 2-D area with a lot of flavor. We could settle for this thin slice, but we want more. With the cake in front of us, we rotate our knife around until we sweep out the volume we want, and then we cut out a large piece. That piece has the area of our thin slice but a much larger scrumptious volume for us to enjoy.
A couple other good examples of solids of revolution include party paper bells.
To find the volume of a 3-D solid, we'll use a similar attack method to the one we used to find the area of a 2-D region.
1) Understand the solid whose volume we're finding.
2) Slice the solid and find the approximate volume of each slice. The difference between the two methods, solids with a known cross-section and solids of revolution, involves slicing.
3) Add up the volumes and take the limit as the number of slices approaches ∞ to get an integral.
All the assumptions we made for finding area still apply if we replace the word "area" with the word "volume." Well, except the first one–when we get into three dimensions we don't really have "vertical" and "horizontal" slices anymore.
We already mentioned that slicing a solid with known cross-section is like slicing a loaf of cinnamon raisin bread. We could also think of it like we're slicing carrots for pot roast or potato slices for potatoes au gratin. Hungry for more?
When asked to find the volume of one of these solids, we're given a few things to start. We'll be told what the base of the solid is, which way it's being sliced,
and what the slices look like.
We can call this one the stick of butter example. Let R be the region enclosed by the x-axis, the graph y = x2, and the line x = 4. Write an integral expression for the volume of the solid whose base is Rand whose slices perpendicular to the x-axis are squares.
Let's use our attack plan.
1) First we have to understand what the solid is.
The region R looks like this:
Let's turn R on its side to make it easier to think of as the base of the solid.
We're told that if the solid is sliced perpendicularly to the x-axis the slices are squares. Since the base of the solid stretches from the x-axis up to the graph y = x2, the side-length of the slice at position x is y = x2. This means if we take slices near x = 0, they'll be tiny. If we take slices near x = 4, they'll be much bigger.
2) Now that we understand what the solid looks like, we need to slice it and find the approximate volume of a slice. We've already sliced it, and we know that each slice is a square with side-length y = x2. Each square has a tiny little bit of thickness Δ x.
To find the volume of the slice we multiply the area of the square by the thickness of the slice to get
(x2)2 Δ x.
3) The variable x goes from 0 to 4 within this solid. When we add the volumes of all the slices and take the limit as the number of slices approaches ∞, we find the volume of the solid is
We recommend drawing pictures. Lots of them. Don't be stingy. You'll use less paper and time drawing a couple extra images than you would by getting the wrong answer and having to start all over. At a minimum, three of them:
1) the region that forms the base of the solid.
2) the region with at least one slice sitting on it.
3) a slice all by itself.
The best way to get better at these is to practice. Feel free to have your favorite 3-D sweet treat while you go through these exercises.
Welcome to the jungle. We won't find any scrumptious sweets here, unless our jungle is in the middle of Oz. Our well-being depends on us being able to tackle finding volumes of more complicated objects, including pyramids, cones, and spheres.
If we just stand here helplessly, we're vulnerable to an attack by a swarm of killer bees. We're going to jump right in to our voyage and see if we can integrate some of these shapes.
Find an integral expression for the volume of a pyramid whose height is 7 and whose base is square with sides of length 5.
The pyramid looks like:
If we slice the pyramid horizontally, we get a square slice:
We can use h as the variable of integration, where h measures the distance from the top of the pyramid to the slice. Then the thickness of the slice is Δ h. To find the side-length of the square, we can cut the pyramid straight down the middle from top to bottom and behold the similar triangles.
Let's have x be the length of the side of the square. Using similar triangles,
The area of the square is
so the volume of the slice at height h is
To find the volume, just add up the volumes of all those slices with an integral
To find the volume of a cone, we do the same thing except that the slices are now circles instead of squares. We use similar triangles to find the radius of each slice. The only difference is that we'd rather eat ice cream out of a cone instead of a pyramid.
Finally we come to the spheres. These are shaped like that baseball your threw at your brother's head when you were five. All these years later, he's still mad about it. And this calculus section is karma's way of repaying the favor.
It turns out that spheres just aren't that complicated. Whether we slice a sphere horizontally or vertically we get circular slices.
First, we need an equation for these circles. The equation of a sphere centered at the origin is
x2 + y2 + z2 = r2.
If we set any one of those variables to 0, the remaining variables satisfy the equation for a circle. For example, if z = 0 then
x2 + y2 = r2.