For simple geometric shapes like rectangles and spheres, it’s no hard task to compute their volumes since clear, simple formulas for volumes of them can be directly applied without further calculation to compute their volumes. The simple geometric shapes are only special cases since if they are put in a chosen coordinate, one has to satisfy very specific relationship between variables in order to maintain such shapes. For more complex and general shapes, they represent most of the shapes occur among all shapes as the graphs of all possible relationships between the variables of the coordinate one chose. Therefore, for more general relationships between variables, namely, for example, in a x-y-z 3-D rectangle coordinate, the relationships between x, y, z, their graphs are more general shapes, and need more generalized ways to calculate their volumes.
Let E be a solid in 3-D space with volume V. For the convenience of using our study methods, we shall put E within a rectangle coordinate in space. One can surly argue that the validity and convenience of using alternative coordinate system for appropriate shapes of solid, yet since E is some solid in general and thus arbitrary, unknown shape, rectangle coordinate has better chance to fit for it. In order to calculate the volume of the solid in general shape, providing only the formulas for simple geometric shapes, one solution is to break down the solid into basic pieces in simple shapes and sum every pieces up. We know that a surface in space can be expressed numerically by some function z=f(x, y), where z is a function of x and y; according to this statement, since a solid is a closed space extending to three degrees of freedom, thus the numerically expression of a solid cannot be one function, since for each set values of x and y, there may be more than 2 corresponding values of z. Therefore, we shall conclude that a solid need to be represented

by at least 2 functions z=f(x, y) and z=g(x, y), providing that in directions of all three degrees of freedom, changes occur somewhere on the solid.
Since E can be expressed with some set of functions z=f(x, y) and z=g(x, y), for each value of z within the range of z, there’s a relationship between variables x and y, in forms of c=f(x, y) or k=g(x, y), where c is within the range of z in f(x, y) and k is within the range of z in g(x, y). With such relationship of x and y for some constant value of z, it creates a region/plane
R in space. For every value of z, there exist a such plane. We can imagine taking constant values of z and getting the corresponding plane as slicing the solid in the direction normal to the z-axis. As the difference between the values taken for z getting smaller, the sum of all slices approaches closer and closer to the shape of the original volume; when the number of slice approaches infinitely close to infinity, the sum of all slices is the solid. The same approximation can be applied in terms of calculating its volume. That is, 𝑉 ≈ ∑ 𝑛 𝑖=1
∆𝑧 ∗ 𝐴𝑟𝑒𝑎(𝑅 𝑖
) , where R i
is the (i) th region/plane corresponding to some value of z (within the range of z). When n approaches infinity, the volume is then 𝑉 = ∫ 𝑧 𝑚𝑎𝑥 𝑧 𝑚𝑖𝑛
𝐴(𝑅 𝑖
) 𝑑𝑧 . With the method for calculating area of regions introduced by double integral, where it tells us that 𝐴(𝑅) = ∬ 1𝑑𝐴 .
Therefore, we can then rewrite the integral of volume as 𝑉 =
∫ 𝑧 𝑚𝑎𝑥 𝑧 𝑚𝑖𝑛
(∬ 1𝑑𝐴 ) 𝑑𝑧 . To deduce the integration limits of the inner double integral, we shall then look at the shape of some region R. A region R is generated when z is set to be some constant, where the constant is within the range of z. Since E is the graph of two function f(x, y) and g(x, y), thus such region R is f(x, y)=c and g(x, y)=k, where c and k are values of z within their function’s range for z. For equations f(x, y)=c and g(x, y)=k, since they are relationships between x and y, then we are able to solve for x or y to be functions of the other variable. By interpreting the graph of such region, we can see that the boundary of such regions cannot be expressed by one function between x and y, since we can see places where multiple values of the dependent variables correspond to one value of the independent variable.
(Surly that we shall switch the dependent and the independent variable around for our purposes, yet for either cases, regardless of which variable between x and y being the dependent or independent variable, there may be
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places where R cannot be expressed by one function, and thus the above statement still holds true.) We can choose y to be the dependent variable and x to be the independent variable, and thus y is some function of x. Note the choice of variable is not a significant factor in this problem since switching them makes no difference in our later conclusion. Then for each region R created by letting z being some constant, the region can be described by two
(or more) functions y=m(x) and y=h(x). For study convenience, we let m(x) > h(x). Note again that the choice of names of such functions are trivial, since one can simply can m(x) to be h(x) and the other way around.
Therefore, we shall now set up the double integral for calculating the area of
R. Let the domain of x for R be [a, b]; then the area of R 𝐴(𝑅) = 𝑏
∫ ∫ 𝑚(𝑥) ℎ(𝑥)
1𝑑𝑦 𝑑𝑥 . Plugging the double integral back into the triple integral that calculates the volume of E, the integral is in form of 𝑉 =
∫ 𝑧 𝑚𝑎𝑥 𝑧 𝑚𝑖𝑛 𝑏
∫ ∫ 𝑚(𝑥) ℎ(𝑥)
1𝑑𝑦 𝑑𝑥 𝑑𝑧 .
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