- Topics At a Glance
- Area
- Assumptions When Finding Area
- Triangles
- Circles
- Integrating with Polar Coordinates
**Volume**- Volumes of Solids with Known Cross-Sections
**Pyramids, Cones, and Spheres**- Volume of Solids of Revolution
- Disks and Washers
- Shells
- Washers vs. Shells
- Arc Length
- Arc Length for Parametric Functions
- Arc Length for Polar Functions
- In the Real World
- I Like Abstract Stuff; Why Should I Care?
- How to Solve a Math Problem

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 helpless, we are vulnerable to an attack by a swarm of killer bees. We are 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.

Answer.

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,

so.

The area of the square is

so the volume of the slice at height *h* is

The volume of the pyramid is

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. This is 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

*x*^{2} + *y*^{2} + *z*^{2} = *r*^{2}.

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

*x*^{2} + *y*^{2} = *r*^{2}.

Example 1

Write an integral expression for the volume of a cone with height 8 and base radius 5. |

Example 2

Now that we have an equation for a circle, we want to write an integral expression for the volume of a sphere with radius 5. |

Exercise 1

Find an integral expression for the volume of a pyramid with height 9 and a square base with side-length 6. Use *h* as the variable of integration, where *h* measures the distance from the base of the pyramid to a slice.

Exercise 2

Find an integral expression for the volume of a pyramid with height *h* and a square base with side-length *b*. Evaluate the integral. This will give you a formula for the volume of a pyramid.

Exercise 3

Write an integral expression for the volume of an inverted (upside-down) cone with height 5 and base radius 2. Use <em>h</em> (the distance from the base to a slice) as the variable of integration.

Exercise 4

Write an integral expression for the volume of a cone with height *h* and base radius *r*. Evaluate your expression to get a formula for the volume of a cone.

Exercise 5

Write an integral expression for the volume of a sphere with radius 4. Use horizontal slices and let *h* be the depth of the slice below the top of the sphere.

Exercise 6

Write an integral expression for the volume of a sphere with radius *r*. Evaluate your expression to get a formula for the volume of a sphere.