Solids

25.1 INTRODUCTION
- 25.1.1 Beyond Surfaces: 3D Solids
- 25.1.2 Building Dimensions: Length, Area, and Now Volume
25.2 LECTURE
25.3 EXAMPLES

25.1 INTRODUCTION

25.1.1 Beyond Surfaces: 3D Solids

-dimensional objects are curves and -dimensional objects are regions or surfaces. In dimension , we deal with solids. The simplest solids imaginable are the cube or the spherical ball. Solids in three dimensional space are usually drawn by plotting their boundary surfaces. A solid polyhedron for example is bound by planes. The first figure shows the solid bound by hyperboloids. It is quite a challenge to compute its volume.¹

**Figure 1.** The "Archimedes revenge problem" asks to prove that : , , has .

25.1.2 Building Dimensions: Length, Area, and Now Volume

While curves have length and regions have area, three dimensional solids have volume. We will in the next lecture look at surface area . In this lecture we look at volume .

25.2 LECTURE

25.2.1 From Basic Solids to Triple Integrals

A basic solid in is a bounded region enclosed by finitely many surfaces . A solid is a finite union of such basic solids. We focus here mostly on . A D integral is defined in the same way as a limit of a Riemann sum which for a given integer is defined as The convergence is proven in the same way. The boundary contribution can be neglected in the limit . If is a parametrization of the solid, then

Theorem 1.

**Figure 2.** Solids in are sets which are unions of solids bound by smooth surfaces. The second solid appears in homework 25.3, the last in 25.2.

25.2.2 Computing Volumes with 3D Integrals and Change of Variables

If is constant , then is the volume of the solid . For a cone we can write where is the unit disc. Its volume is . For the unit sphere for example, we can write where is the unit disc . In polar coordinates, we get We can also use spherical coordinates where . The volume is

25.2.3 Two Key Approaches to 3D Integration

There are two basic strategies to compute the integral: the first is to slice the region up along a line like the -axis then form . To get the volume of a cone for example, integrate . The inner double integral is the area of the slice which is . The last integral gives . A second reduction is to see the solid sandwiched between two graphs of a function on a region , then form . In the cone case, we have for the disc of radius . The lower function is the upper function is . We get , a double integral which best can be computed using polar coordinates: Burgers and fries!

**Figure 3.** The "burger and fries methods" to compute triple integral. The first reduces to a single integral, the second to a double integral.

25.2.4 Jacobians for Spherical and Cylindrical Coordinates

We have seen in the theorem the coordinate change formula if is given. For spherical coordinates we have . For cylindrical coordinates, the situation is the same as for polar coordinates. The map produces .

25.2.5 Ellipsoid Volume

Let us find the integral , where is a solid ellipsoid. The most comfortable way is to introduce another coordinate change which maps the solid sphere to to the solid ellipsoid . Then take the spherical coordinate map , where Now is a coordinate change which maps to the ellipsoid. By the chain rule, the distortion factor is . The integral is

25.2.6 Solid Torus Volume: A Special Coordinate System

In order to compute the volume of a solid torus, we can introduce a special coordinate system The solid torus is then the image of the cuboid The determinant is . Integration over the cuboid gives the volume .

25.3 EXAMPLES

Example 1. To find for set up the integral . Start with the core , then integrate the middle layer, and finally handle the outer layer: .

Example 2. To find the moment of inertia of a sphere , we use spherical coordinates. We know that and the distortion factor is . We have therefore We will see some details in class. If we rotate the sphere around the -axis with angular velocity , then is the kinetic energy of that sphere. For example, the moment of inertia of the earth is . With an angular velocity of , this rotational kinetic energy is

Example 3. Problem: Find the volume of the intersection of , and .
Solution: look at ’th of the body given in cylindrical coordinates , . The roof is because above the "one eighth disc" only the cylinder matters. The polar integration problem has an inner -integral of . Integrating this over can be done by integrating by parts (using ) leading to the anti-derivative of . The result is .

Example 4. Problem: A pencil , a hexagonal cylinder of radius above the -plane is cut by a sharpener below the cone . What is its volume?
Solution: we consider one sixth of the pen where the base is the polar region and . The pen’s back is and the sharpened part is . The integral can be computed and is a bit messy .²

The homework for this unit and the next unit is combined and will be given at the end of the next one.

Archimedes Revenge, first appeared in Math S21a exam, Harvard Summer School, 2017.↩︎
An exam problem at ETH in a single variable calculus exam when Oliver was an undergrad.↩︎

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