# Calculus/Multiple integration
## Double integration

For (Riemann) integrals, we consider the Riemann sum. Recall in the one-variable case, we partition an interval into more and more subintervals with smaller and smaller width, and we are integrating over the interval by summing the areas of corresponding rectangles for each subinterval. For the multivariable case, we need to do something similar, but the problem arises when we need to partition 'interval' in $\mathbb {R} ^{2},\mathbb {R} ^{3}$ or $\mathbb {R} ^{n}$ in general. (Actually, we only have the term interval in $\mathbb {R}$.)
In multivariable case, we need to consider not just 'interval' itself (which is undefined in multivariable case), but Cartesian product over intervals for $\mathbb {R} ^{2}$, and more generally n-ary Cartesian product over intervals for $\mathbb {R} ^{n}$.
Definition. (n-ary Cartesian product) The n-ary Cartesian product over $n$ sets $X_{1},X_{2},\ldots ,X_{n}$ is the set $X_{1}\times X_{2}\times \cdots \times X_{n}=\{(x_{1},x_{2},\ldots ,x_{n}):x_{i}\in X_{i}{\text{ for each }}i\in \{1,2,\ldots ,n\}\}$ of $n$-tuples (or vectors).
Remark.
* recall that interval is essentially a set, e.g. $[0,1]=\left\{x:0\leq x\leq 1\right\}$.
* 2-ary Cartesian product is simply called Cartesian product, e.g. $[0,1]\times [0,2]=\{(x_{1},x_{2}):x_{1}\in [0,1]{\text{ and }}x_{2}\in [0,2]\}$ is Cartesian product over two intervals, and also a rectangle with side lengths 1 and 2 geometrically.
* special case: $\underbrace {X\times X\times \cdots X} _{n\,X's}$ is called n-ary Cartesian power and is denoted by $X^{n}$, e.g. $\mathbb {R} ^{3}=\mathbb {R} \times \mathbb {R} \times \mathbb {R} =\{(x,y,z):x\in \mathbb {R} ,y\in \mathbb {R} {\text{ and }}z\in \mathbb {R} \}$.
Area (for $n=2$), volume (for $n=3$) or measure (for each positive number $n$) of geometric objects (e.g. rectangles in $\mathbb {R} ^{2}$ and cubes in $\mathbb {R} ^{3}$) in $\mathbb {R} ^{n}$ is the product of the lengths of all its sides (in different dimensions).
Example.
* $[1,2]\times [3,4]$ is the Cartesian product over two intervals (It is square with side length $1$ in $\mathbb {R} ^{2}$ geometrically)
* $[1,2]\times [{\sqrt {2}},10]\times [-\pi ,-1]$ is the 3-ary Cartesian product over three intervals (It is a rectangular cuboid in $\mathbb {R} ^{3}$ geometrically)
* $[0.3,1.7]\times [0.3,1.7]\times [0.3,1.7]\times [0.3,1.7]$ is the 4-ary Cartesian product over four intervals, or 4-ary Cartesian power (it is denoted as $[0.3.1.7]^{4}$) (It is a 4-dimensional cube in $\mathbb {R} ^{4}$ geometrically)
Now, we are ready to define multiple integral in an analogous way compared with single integral. For simplicity, let us first discuss double integral, and then generalize it to multiple integral in an analogous way.
Definition. (Double integrals) Let $f(x,y)$ be a function defined on a rectangle $R$ in $\mathbb {R} ^{2}$. Consider a partition of $R$ into small rectangles with areas $\Delta A_{1},\Delta A_{2},\ldots ,\Delta A_{n}$ respectively. Choose an arbitrary point $(x_{k},y_{k})$ in the $k$th rectangle. The function $f$ is integrable over $R$ if $\lim _{\Delta A_{k}\to 0}\sum _{k=1}^{n}f(x_{k},y_{k})\Delta A_{k}$ exists. In that case, we denote this limit by $\iint _{R}f\,dA$ ($A$ is a mnemonic of area), and call it the double integral of $f$ over $R$.
Remark.
* the limit $\lim _{\Delta A_{k}\to 0}\sum _{k=1}^{n}f(x_{k},y_{k})\Delta A_{k}$ is the Riemann sum over $R$ with the partition of $R$ into small rectangles
* as we partition more and more small rectangles (with smaller and smaller area) from the rectangle $R$, which arbitrary point we choose in each rectangle becomes less and less important, because different points in each small rectangle become 'closer and closer' together, and thus the position of different points are more and more 'similar'
* we may assume that nice functions (functions given in the questions here are nice, unless stated otherwise) to be integrable, so we do not need to check integrability of every function we encounter here (Checking of integrability involves the formal definition of limit, and is out of scope here)
* the process of computing double integral is called double integration
A physical meaning of double integration is computing volume.
Proposition. (Volume given by double integration) Let $f(x,y)$ be an integrable function defined on a rectangle $R$ in $\mathbb {R} ^{2}$. Suppose $f(x,y)\geq 0$ for each $(x,y)\in R$. Then, the volume under the graph of $f$ over $R$ is $\iint _{R}f\,dA.$
Remark.
* if $f(x,y)$ is not always $\geq 0$, the region below the $xy$-plane has negative volume, so this negative volume will cancel out the positive volume, which may or may not be desired
* if $f(x,y)$ is always negative, then the volume computed by this formula is negative, and we usually take its absolute value to get the volume, because volume is usually defined to be nonnegative
Let's also introduce some properties of double integral to ease computation of double integral.
Proposition. (Properties of double integral) Let $f(x,y)$ and $g(x,y)$ be integrable functions defined on a rectangle $R$ in $\mathbb {R} ^{2}$. Then, the following properties hold.
* (Linearity) $af+bg$ is also integrable over $R$ for each real number $a$ and $b$, and $\iint _{R}(af+bg)\,dA=a\iint _{R}f\,dA+b\iint _{R}g\,dA$
* (Monotonicity) if $f(x,y)\leq g(x,y)$ for each $(x,y)\in R$, then $\iint _{R}f\,dA\leq \iint _{R}g\,dA$
* (Triangle inequality) $|f|$ is integrable over $R$ and $\left|\iint _{R}f\,dA\right|\leq \iint _{R}|f|\,dA$
Proposition. (Continuity implies integrability) Let $f(x,y)$ be a continuous function defined on a rectangle $R$ in $\mathbb {R} ^{2}$. Then, the function $f$ is integrable.
Remark. Because nice functions are often continuous, most nice functions are integrable.

### Iterated integrals

Thankfully, we need not always work with Riemann sums every time we want to calculate an integral in more than one variable. There are some results that make life a bit easier for us. Before stating the result, we need to define iterated integral, which is used in the results.
Definition. (Iterated integral) Let $f(x,y)$ be a continuous function defined on a rectangle $[a,b]\times [c,d]$ in $\mathbb {R} ^{2}$. The iterated integral is defined by $\int _{a}^{b}\int _{c}^{d}f(x,y)\,dy\,dx=\int _{a}^{b}\left(\int _{c}^{d}f(x,y)\,dy\right)dx$ and $\int _{c}^{d}\int _{a}^{b}f(x,y)\,dx\,dy=\int _{c}^{d}\left(\int _{a}^{b}f(x,y)\,dx\right)dy$
Remark. e.g., for the integral $\int _{c}^{d}\left(\int _{a}^{b}f(x,y)\,dx\right)dy$, we first compute the definite integral $\int _{c}^{d}f(x,y)\,dy$ with respect to $y$ by treating $x$ as a constant. Then, we compute the remaining integral (in terms of $x$) with respect to $x$.
Computation of iterated integrals is generally much easier than computing the double integral directly using Riemann sum. So, it will be nice if we have some relationships between iterated integral and double integral for us to compute double integral with the help of iterated integral. It is indeed the case and the following theorem is the bridge between iterated integral and double integral.
Theorem. (Fubini's theorem) Let $f(x,y)$ be a continuous function defined on a rectangle $R=[a,b]\times [c,d]$ in $\mathbb {R} ^{2}$. Then, $\iint _{R}f\,dA=\int _{a}^{b}\int _{c}^{d}f(x,y)\,dy\,dx=\int _{c}^{d}\int _{a}^{b}f(x,y)\,dx\,dy.$
Remark.
* i.e., we can use iterated integral in either order to compute the corresponding double integral
* we should notice the change of bounds for each integral after changing the integration order
Example. (Volume of rectangular cuboid)
(i) Prove that the volume under the graph of $z=4$ over the rectangle $[0,2]\times [0,3]$ is $24$.
(ii) Hence, prove that volume of rectangular cuboid with length $\ell$, width $w$ and height $h$ is $\ell wh$.
(i) Because $z=4>0$ for each $(x,y)\in [0,2]\times [0,3]$, by the proposition of volume given by double integration, the volume is $\iint _{[0,2]\times [0,3]}4\,\,dA$. By Fubini's theorem, this equals $\int _{0}^{2}\int _{0}^{3}4\,dy\,dx=4\int _{0}^{2}\int _{0}^{3}\,dy\,dx=4\int _{0}^{2}(3-0)\,dx=4(3(2)-3(0))=24.$
Remark. Geometrically, the graph is a rectangular cuboid, and its volume is given by product of each side length, namely $2(3)(4)=24$, which matches with our answer.
(ii) The desired volume is given by $\iint _{[0,\ell ]\times [0,w]}h\,dx\,dy$ (double integral of a constant function $z=h$ over the rectangle $[0,\ell ]\times [0,w]$ with length $\ell$ and width $w$). (We may also express the integral as $\iint _{[0,\ell ]\times [0,w]}h\,dy\,dx$ without affecting the result.) Then, by Fubini's theorem, $\iint _{[0,\ell ]\times [0,w]}h\,dx\,dy=\int _{0}^{\ell }\int _{0}^{w}h\,dy\,dx=h\int _{0}^{\ell }\int _{0}^{w}\,dy\,dx=h\int _{0}^{\ell }w\,dx=\ell wh.$ $\Box$

### Double integrals over more general regions in R$^2$

We have defined double integrals over rectangles in $\mathbb {R} ^{2}$. However, we often want to compute double integral over regions with shape other than rectangle, e.g. circle, triangle, etc. Therefore, we will discuss an approach to compute double integrals over more general regions reasonably, without altering the definition of double integrals.
Consider a function $f:D\to \mathbb {R}$ in which $D\subseteq \mathbb {R} ^{2}$ is a general region. To apply the definition of double integrals, we need to transform the general region $D$ to a rectangle (say $R$). An approach is finding a rectangle $R$ containing $D$ (i.e., $R\supseteq D$), and let $f(x,y)=0$ for each $(x,y)\in R$ lying outside $D$ (i.e., for each $(x,y)\in R\setminus D$ ). Because the value of the function is zero outside the region we are integrating over, this does not change the volume under the graph of $f$ over $D$, so this way is a good way to define such double integrals. Let's define such double integrals formally in the following.
Definition. (Double integrals over general regions) Let $f(x,y)$ be a function defined on a region (of arbitrary shape) $D\subseteq R\subseteq \mathbb {R} ^{2}$. Then, we define $f(x,y)=0$ for each $(x,y)\in R$ lying outside $D$ (i.e., for each $(x,y)\in R\setminus D$ ), and define the double integral of function $f$ over the region $D$ by $\iint _{D}f\,dA=\iint _{R}f\,dA$ if the latter integral exists.
Remark. Then, we may compute double integrals over general regions by computing the corresponding Riemann sum for the latter integral.
However, this way of computation (by computing Riemann sum) is generally very difficult, and usually we use a generalized version of Fubini's theorem to compute such integrals. It will be discussed in the following.
Theorem. (Generalized Fubini's theorem) Let $f(x,y)$ be a continuous function defined on a region (with arbitrary shape) $D$. Then, the following hold.
(i) If $D=\{(x,y):a\leq x\leq b{\text{ and }}g_{1}(x)\leq y\leq h_{1}(x)\}$ in which functions $g_{1}$ and $h_{1}$ are continuous, then $\iint _{D}f\,dA=\int _{a}^{b}\int _{g_{1}(x)}^{h_{1}(x)}f(x,y)\,dy\,dx.$ (ii) If $D=\{(x,y):c\leq y\leq d{\text{ and }}g_{2}(y)\leq x\leq h_{2}(y)\}$ in which functions $g_{2}$ and $h_{2}$ are continuous, then $\iint _{D}f\,dA=\int _{c}^{d}\int _{g_{2}(y)}^{h_{2}(y)}f(x,y)\,dx\,dy.$ in which $a,b,c,d$ are real numbers satisfying the above conditions.
Proof. We can prove this theorem by Fubini's theorem (ungeneralized version) and applying the definition of double integrals over general regions. (We can use Fubini's theorem because we assume that functions $g$ and $h$ are continuous.)
Part (i): ($D=\{(x,y):a\leq x\leq b{\text{ and }}g(x)\leq y\leq h(x)\}$)
Take arbitrary rectangle $R=[a,b]\times [c,d]$ containing $D$ (i.e. $c\leq g(x)$ and $d\geq h(x)$ ). Then, define $f(x,y)=0$ if $(x,y)\in R\setminus D$. After that, $\iint _{D}f\,dA{\overset {\text{ def }}{=}}\iint _{R}f\,dA=\int _{a}^{b}\int _{c}^{d}f(x,y)\,dy\,dx=\int _{a}^{b}\left(\underbrace {\int _{c}^{g(x)}f(x,y)\,dy} _{0{\text{ because }}y<({\text{ or }}\leq )g(x)}+\int _{g(x)}^{h(x)}f(x,y)\,dy+\underbrace {\int _{h(x)}^{d}f(x,y)\,dy} _{0{\text{ because }}y>({\text{ or }}\geq )h(x)}\right)dx.$ The result follows. (We say that whether equality holds in the above inequalities does not matter, because the definite integral over a point equals zero anyway, so it does not affect the result.)
Part (ii): ($D=\{(x,y):c\leq y\leq d{\text{ and }}g(x)\leq x\leq h(x)\}$)
Similarly, take arbitrary rectangle $R=[a,b]\times [c,d]$ containing $D$ (i.e. $a\leq g(x)$ and $b\geq h(x)$ ). Then, define $f(x,y)=0$ if $(x,y)\in R\setminus D$. After that, $\iint _{D}f\,dA{\overset {\text{ def }}{=}}\iint _{R}f\,dA=\int _{c}^{d}\int _{a}^{b}f(x,y)\,dx\,dy=\int _{c}^{d}\left(\underbrace {\int _{a}^{g(x)}f(x,y)\,dx} _{0{\text{ because }}x<({\text{ or }}\leq )g(x)}+\int _{g(x)}^{h(x)}f(x,y)\,dx+\underbrace {\int _{h(x)}^{b}f(x,y)\,dx} _{0{\text{ because }}x>({\text{ or }}\geq )h(x)}\right)dx.$ The result follows. $\Box$
Remark. Usually, finding the bounds for $x$ and $y$ is the most difficult step when we compute such double integrals.
Example. Let $D$ be the triangle in $\mathbb {R} ^{2}$ with vertices $(0,0),(0,1)$ and $(1,0)$. Prove that $\iint _{D}xy^{2}\,\,dA={\frac {1}{60}}$.
Approach 1: The bound for $x$ is $0\leq x\leq 1$. Given a fixed $x$, the bound for $y$ is $0\leq y\leq 1-x$. Thus, the integral is ${\begin{aligned}\int _{0}^{1}\int _{0}^{1-x}xy^{2}\,dy\,dx&=\int _{0}^{1}\left[{\frac {1}{3}}y^{3}x\right]_{y=0}^{y=1-x}\,dx\\&={\frac {1}{3}}\int _{0}^{1}x(1-x)^{3}\,dx\\&={\frac {1}{3}}\int _{0}^{1}\left(x-3x^{2}+3x^{3}-x^{4}\right)\,dx\\&={\frac {1}{3}}\left({\frac {1}{2}}-1+{\frac {3}{4}}-{\frac {1}{5}}\right)\\&={\frac {1}{60}}.\end{aligned}}$ Approach 2: The bound for $y$ is $0\leq y\leq 1$. Given a fixed $y$, the bound for $x$ is $0\leq x\leq 1-y$. Thus, the integral is ${\begin{aligned}\int _{0}^{1}\int _{0}^{1-y}xy^{2}\,dx\,dy&=\int _{0}^{1}\left[{\frac {1}{2}}x^{2}y^{2}\right]_{x=0}^{x=1-y}\,dy\\&=\int _{0}^{1}{\frac {1}{2}}(1-y)^{2}y^{2}\,dy\\&={\frac {1}{2}}\int _{0}^{1}\left(y^{2}-2y^{3}+y^{4}\right)\,dy\\&={\frac {1}{2}}\left({\frac {1}{3}}-{\frac {2}{4}}+{\frac {1}{5}}\right)\\&={\frac {1}{60}}.\end{aligned}}$ $\Box$
Example. (Volume of tetrahedron) Consider a tetrahedron in $\mathbb {R} ^{3}$ with vertices $(0,0,0),(a,0,0),(0,b,0)$ and $(0,0,c)$ in which $a,b$ and $c$ are positive numbers. Define vectors $\mathbf {a} =(a,0,0),\mathbf {b} =(0,b,0)$ and $\mathbf {c} =(0,0,c)$. Prove that volume of the tetrahedron is ${\frac {1}{6}}(\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} ))$. (i.e., ${\frac {1}{6}}$ of the volume of parallelepiped spanned by the vectors $\mathbf {a} ,\mathbf {b}$ and $\mathbf {c}$)
Let the plane containing $(a,0,0),(0,b,0)$ and $(0,0,c)$ be $\Pi$. To find the equation of $\Pi$, consider its normal vector. A normal vector of $\Pi$ is $[(a,0,0)-(0,0,c)]\times [(0,b,0)-(0,0,c)]=(a,0,-c)\times (0,b,-c)={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\a&0&-c\\0&b&-c\end{vmatrix}}=(bc,ac,ab).$ Therefore, the equation of $\Pi$ is ${\begin{aligned}&&bc(x-a)+ac(y-0)+ab(z-0)&=0\\&\Rightarrow &bcx-abc+acy+abz&=0\\&\Rightarrow &{\frac {bcx}{abc}}-{\frac {abc}{abc}}+{\frac {acy}{abc}}+{\frac {abz}{abc}}&=0\\&\Rightarrow &{\frac {x}{a}}+{\frac {y}{b}}+{\frac {z}{c}}&=1\\&\Rightarrow &z&=c\left(1-{\frac {x}{a}}-{\frac {y}{b}}\right).\end{aligned}}$ The desired volume is the volume under the graph of $f(x,y)=c\left(1-{\frac {x}{a}}-{\frac {y}{b}}\right)$ over a region $D$, and the region $D$ is the projection of the tetrahedron on the xy-plane, which is the triangle with vertices $(0,0),(a,0),(0,b)$ in $\mathbb {R} ^{2}$. Because the line passing through $(a,0)$ and $(0,b)$ has equation $y=b\left(1-{\frac {x}{a}}\right)$, the bound for $x$ is $0\leq x\leq a$, and the bound for $y$ is $0\leq y\leq b\left(1-{\frac {x}{a}}\right)$ given a fixed $x$. Thus, the desired volume is ${\begin{aligned}\int _{0}^{a}\int _{0}^{b\left(1-{\frac {x}{a}}\right)}\underbrace {c\left(1-{\frac {x}{a}}-{\frac {y}{b}}\right)} _{{\text{Equation of }}\Pi }\,dy\,dx&=c\int _{0}^{a}\left(\left(1-{\frac {x}{a}}\right)\left(b\left(1-{\frac {x}{a}}\right)\right)-{\frac {b^{2}\left(1-{\frac {x}{a}}\right)^{2}}{2b}}\right)\,dx\\&=c\int _{0}^{a}\left(b\left(1-{\frac {x}{a}}\right)^{2}-{\frac {b}{2}}\left(1-{\frac {x}{a}}\right)^{2}\right)\,dx\\&=c\int _{0}^{a}\left({\frac {b}{2}}\left(1-{\frac {x}{a}}\right)^{2}\right)\,dx\\&={\frac {bc}{2}}\int _{0}^{a}\left(1-{\frac {2x}{a}}+{\frac {x^{2}}{a^{2}}}\right)\,dx\\&={\frac {bc}{2}}\left(a-{\frac {a^{2}}{a}}+{\frac {a^{3}}{3a^{2}}}\right)\\&={\frac {bc}{2}}\left(a-a+{\frac {a}{3}}\right)\\&={\frac {abc}{6}}.\end{aligned}}$ On the other hand, ${\frac {1}{6}}(\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} ))={\frac {1}{6}}\left((a,0,0)\cdot {\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\0&b&0\\0&0&c\end{vmatrix}}\right)={\frac {1}{6}}((a,0,0)\cdot (bc,0,0))={\frac {abc}{6}}$ which equals the desired volume. $\Box$
Example. (Switching integration order) Prove that $\int _{0}^{16}\int _{x^{1/4}}^{2}{\frac {1}{1+y^{5}}}\,dy\,dx$ is approximately $0.70$. (correct to two decimal places)
If we integrate in this order, the computation will be very tedious. Thus, we will interchange the integration order using generalized Fubini's theorem to ease the computation. Originally, the bounds are $0\leq x\leq 16$ and $x^{1/4}\leq y\leq 2$ (given a fixed $x$). If we integrate over $x$ first instead, the suitable expressions of bounds are $0\leq y\leq 2$ and $0\leq x\leq y^{4}$ (given a fixed $y$). Therefore, ${\begin{aligned}\int _{0}^{16}\int _{x^{1/4}}^{2}{\frac {1}{1+y^{5}}}\,dy\,dx&=\int _{0}^{2}\int _{0}^{y^{4}}{\frac {1}{1+y^{5}}}\,dx\,dy\\&=\int _{0}^{2}{\frac {y^{4}}{1+y^{5}}}\,dy\\&={\frac {1}{5}}\int _{\underbrace {1} _{1+0^{5}}}^{\overbrace {33} ^{1+2^{5}}}{\frac {1}{u}}\,du\qquad ({\text{let }}u=1+y^{5}\implies du=5y^{4}\,dy)\\&={\frac {1}{5}}(\ln 33-\underbrace {\ln 1} _{0})\approx 0.70.\end{aligned}}$
Proposition. (Union of regions for double integration) Let $f(x,y)$ be a continuous function defined on a bounded set $\color {green}{D}$ in $\mathbb {R} ^{2}$. If $\color {green}{D}$ is the union of $\color {green}{D_{1}}$ and $\color {green}{D_{2}}$, in which the overlapping set of $\color {green}{D_{1}}$ and $\color {green}{D_{2}}$ (in $\mathbb {R} ^{2}$) has zero area, then $\iint _{\color {green}D}f\,dA=\iint _{\color {green}{D_{1}}}f\,dA+\iint _{\color {green}{D_{2}}}f\,dA$ if the integrals exist.
Remark.
* Example of analogous case in $\mathbb {R}$: $\int _{[0,2]}fdx=\int _{[0,1]}fdx+\int _{[1,2]}fdx$ (the overlapping set of $[0,1]$ and $[1,2]$ has zero length)
* Curves and points have zero area in $\mathbb {R} ^{2}$.
Corollary. (Subtraction of regions for double integration) Let $f(x,y)$ be a continuous function defined on a bounded set $\color {green}{D}$ in $\mathbb {R} ^{2}$. If $\color {green}{D}$ is the union of $\color {red}{D_{1}}$ and $\color {blue}{D_{2}}$, in which ${\color {green}{D}}={\color {blue}{D_{2}}}\setminus {\color {red}{D_{1}}}$,then $\iint _{\color {green}{D}}f\,dA=\iint _{\color {blue}{D_{2}}}f\,dA-\iint _{\color {red}{D_{1}}}f\,dA$ if the integrals exist.
Proof. First, the overlapping set of ${\color {green}D}$ and ${\color {red}D_{1}}$ is ${\color {green}D}\cap {\color {red}D_{1}}=({\color {blue}D_{2}}\setminus {\color {red}D_{1}})\cap {\color {red}D_{1}}={\color {blue}D_{2}}\cap \overbrace {{\color {red}D_{1}}^{c}\cap {\color {red}D_{1}}} ^{\varnothing }=\varnothing$, which has zero area. Next, because ${\color {green}D}={\color {blue}D_{2}}\setminus {\color {red}D_{1}}\implies {\color {green}D}={\color {blue}D_{2}}\cap {\color {red}D_{1}}^{c}\implies {\color {green}D}\cup {\color {red}D_{1}}={\color {blue}D_{2}}\cap {\color {red}D_{1}}^{c}\cup {\color {red}D_{1}}\implies {\color {blue}D_{2}}={\color {green}D}\cup {\color {red}D_{1}}$, $\iint _{\color {blue}D_{2}}f\,dA=\iint _{\color {green}D}f\,dA+\iint _{\color {red}D_{1}}f\,dA.$ The result follows. $\Box$
Proposition. (Area given by double integration) Let $D$ be a bounded region in $\mathbb {R} ^{2}$. Then, the area of $D$ is $\iint _{D}1\,dA.$
Proof. The volume under the graph of the constant function $f\equiv 1$ over $D$ equals the base area (which is the area of $D$) times the height (the height is one in this case). Therefore, the area of $D$ is the volume under the graph of $f$ over $D$, which is $\iint _{D}f\,\,dA=\iint _{D}1\,\,dA$ by proposition about volume given by double integration (because $1>0$ for each $(x,y)\in D$ ) $\Box$
Remark. Recall that area of a bounded region can also be found by single integration. In some cases, using this proposition instead is more convenient.
Example. Let $D$ be a region bounded by the curves $y={\sqrt {r^{2}-x^{2}}}$ and $y=-{\sqrt {r^{2}-x^{2}}}$, in which $r$ is a positive number. Prove that the area of bounded region $D$ is $\pi r^{2}$.
Solving $y={\sqrt {r^{2}-x^{2}}}$ and $y=-{\sqrt {r^{2}-x^{2}}}$, we get ${\sqrt {r^{2}-x^{2}}}=-{\sqrt {r^{2}-x^{2}}}\implies {\sqrt {r^{2}-x^{2}}}=0\implies x=\pm r$. Because $x=\pm r\implies y=0$, intersection points of these two curves are $({\sqrt {r}},0)$ and $(-{\sqrt {r}},0)$. Therefore, the bound for $x$ is $-{\sqrt {r}}\leq x\leq {\sqrt {r}}$, and given a fixed $x$, the bound for $y$ is $-{\sqrt {r^{2}-x^{2}}}\leq y\leq {\sqrt {r^{2}-x^{2}}}$. Thus, the desired area is ${\begin{aligned}\iint _{D}1\,dA&=\int _{-r}^{r}\int _{-{\sqrt {r^{2}-x^{2}}}}^{\sqrt {r^{2}-x^{2}}}1\,dy\,dx\\&=2\int _{-r}^{r}{\sqrt {r^{2}-x^{2}}}\,dx\\&=2\int _{\arcsin(-r/r)}^{\arcsin(r/r)}r\cos \theta {\sqrt {r^{2}-r^{2}\sin ^{2}\theta }}\,d\theta \qquad {\text{let }}x=r\sin \theta \implies dx=r\cos \theta \,d\theta {\text{ and }}\theta =\arcsin(x/r)\\&=2r\int _{\underbrace {\arcsin(-1)} _{-\pi /2}}^{\overbrace {\arcsin 1} ^{\pi /2}}\cos \theta \underbrace {\sqrt {r^{2}}} _{=|r|=r{\text{ because }}r>0}\cos \theta \,d\theta \\&=2r^{2}\int _{-\pi /2}^{\pi /2}{\frac {1+\cos(2\theta )}{2}}\,d\theta \qquad {\text{by double angle formula:}}\cos(2\theta )=2\cos ^{2}\theta -1\implies \cos ^{2}\theta ={\frac {1+\cos(2\theta )}{2}}\\&={\frac {2r^{2}}{2}}\left[{\frac {\theta }{2}}+{\frac {\sin 2(\theta )}{4}}\right]\\&={\frac {2r^{2}}{2}}\left({\frac {\pi }{2}}+{\frac {1}{4}}\cdot \underbrace {\sin(2(\pi /2))} _{=\sin \pi =0}-\left(-{\frac {\pi }{2}}+{\frac {1}{4}}\cdot \underbrace {\sin(2(-\pi /2))} _{=\sin(-\pi )=0}\right)\right)\\&=\pi r^{2}\end{aligned}}$ $\Box$ Remark. Geometrically, the bounded region $D$ is a disk (region in a plane bounded by a circle) of radius $r$.

## Triple integration

The concepts in the section of double integrals apply to triple integrals (and also multiple integrals generally) analogously. We will give several examples for triple integrals in this section.
Definition. (Triple integrals) Let $f(x,y,z)$ be a function defined on a rectangular box (or rectangular cuboid) $B$ in $\mathbb {R} ^{3}$. Consider a partition of $B$ into small boxes with volumes $\Delta V_{1},\Delta V_{2},\ldots ,\Delta V_{n}$ respectively. Choose an arbitrary point $(x_{k},y_{k},z_{k})$ in the $k$th box. The function $f$ is integrable over $R$ if $\lim _{\Delta V_{k}\to 0}\sum _{k=1}^{n}f(x_{k},y_{k},z_{k})\Delta V_{k}$ exists. In that case, we denote this limit by $\iiint _{B}f\,dV$ ($V$ is a mnemonic of volume), and call it the triple integral of $f$ over $B$.
Remark.
* The process of computing triple integral is called triple integration
Theorem. (Generalized Fubini's theorem (triple integrals version)) Let $f(x,y,z)$ be a continuous function defined on a region (with arbitrary shape) $D$. Then, the following hold.
(i) if $D=\{(x,y,z):a_{1}\leq x\leq b_{1},g_{1}(x)\leq y\leq h_{1}(x),\varphi _{1}(x,y)\leq z\leq \psi _{1}(x,y)\}$, then $\iiint _{D}fdV=\int _{a_{1}}^{b_{1}}\int _{g_{1}(x)}^{h_{1}(x)}\int _{\varphi _{1}(x,y)}^{\psi _{1}(x,y)}f\,dz\,dy\,dx$ (ii) if $D=\{(x,y,z):a_{2}\leq x\leq b_{2},g_{2}(x)\leq z\leq h_{2}(x),\varphi _{2}(x,z)\leq y\leq \psi _{2}(x,z)\}$, then $\iiint _{D}fdV=\int _{a_{2}}^{b_{2}}\int _{g_{2}(x)}^{h_{2}(x)}\int _{\varphi _{2}(x,z)}^{\psi _{2}(x,z)}f\,dy\,dz\,dx$ (iii) if $D=\{(x,y,z):a_{3}\leq y\leq b_{3},g_{3}(y)\leq x\leq h_{3}(y),\varphi _{3}(x,y)\leq z\leq \psi _{3}(x,y)\}$, then $\iiint _{D}fdV=\int _{a_{3}}^{b_{3}}\int _{g_{3}(y)}^{h_{3}(y)}\int _{\varphi _{3}(x,y)}^{\psi _{3}(x,y)}f\,dz\,dx\,dy$ (iv) if $D=\{(x,y,z):a_{4}\leq y\leq b_{4},g_{4}(y)\leq z\leq h_{4}(y),\varphi _{4}(y,z)\leq x\leq \psi _{4}(y,z)\}$, then $\iiint _{D}fdV=\int _{a_{4}}^{b_{4}}\int _{g_{4}(y)}^{h_{4}(y)}\int _{\varphi _{4}(y,z)}^{\psi _{4}(y,z)}f\,dx\,dz\,dy$ (v) if $D=\{(x,y,z):a_{5}\leq z\leq b_{5},g_{5}(z)\leq y\leq h_{5}(z),\varphi _{5}(y,z)\leq x\leq \psi _{5}(y,z)\}$, then $\iiint _{D}fdV=\int _{a_{5}}^{b_{5}}\int _{g_{5}(z)}^{h_{5}(z)}\int _{\varphi _{5}(y,z)}^{\psi _{5}(y,z)}f\,dx\,dy\,dz$ (vi) if $D=\{(x,y,z):a_{6}\leq z\leq b_{6},g_{6}(z)\leq x\leq h_{6}(z),\varphi _{6}(x,z)\leq y\leq \psi _{6}(x,z)\}$, then $\iiint _{D}fdV=\int _{a_{6}}^{b_{6}}\int _{g_{6}(z)}^{h_{6}(z)}\int _{\varphi _{6}(x,z)}^{\psi _{6}(x,z)}f\,dy\,dx\,dz$ in which each function involved is continuous. That is, we can use either one of all $3!=6$ possible integration orders for iterated integrals to compute triple integrals, with suitable bounds.
Proposition. (4-dimensional volume given by triple integration) Let $f(x,y,z)$ be an integrable function defined on a rectangular box $B$ in $\mathbb {R} ^{3}$. Suppose $f(x,y,z)\geq 0$ for each $(x,y,z)\in B$. Then, the 4-dimensional volume under the graph of $f$ over $B$ is $\iiint _{B}fdV.$
Remark. It is just a theoretical result, and is difficult to be visualized.
Proposition. (Volume given by triple integration) Let $D$ be a bounded region in $\mathbb {R} ^{3}$. Then, the volume of $D$ is $\iiint _{D}1dV.$
Example. Consider a region $D$ that is bounded by a sphere (with radius $r$, which is positive) $x^{2}+y^{2}+z^{2}=r^{2}$ that lies in the octant (+,+,+) (i.e., the octant in which $x,y$ and $z$ are positive).
(i) Prove that the volume of $D$ is ${\frac {1}{6}}\pi r^{3}$.
(ii) Hence, prove that the volume of the region that is bounded by the sphere (in $\mathbb {R} ^{3}$) is ${\frac {4}{3}}\pi r^{3}$.
(i) The given bounds for $D$ is $x^{2}+y^{2}+z^{2}\leq r^{2},x\geq 0,y\geq 0$ and $z\geq 0$. We can express the bounds as follows:
* $x^{2}\leq x^{2}+y^{2}+z^{2}\leq r^{2}{\text{ and }}x\geq 0\implies 0\leq x\leq r$
* given a fixed $x$, aim: finding bounds for $y$ in the form of $g(x)\leq y\leq h(x)$.
   * Steps: $y^{2}\leq y^{2}+\underbrace {z^{2}} _{\geq 0}\underbrace {\leq r^{2}-x^{2}} _{{\text{by }}x^{2}+y^{2}+z^{2}\leq r^{2}}{\text{ and }}y\geq 0\implies 0\leq y\leq {\sqrt {r^{2}-x^{2}}}$
* given fixed $x$ and $y$ ($x$ is selected from its bound, namely $0\leq x\leq r$, and $y$ is selected from its bound, which depends on the fixed $x$), aim: finding bounds for $z$ in the form of $\varphi (x,y)\leq z\leq \psi (x,y)$.
   * Steps: $x^{2}\underbrace {\leq r^{2}-y^{2}-z^{2}} _{{\text{by }}x^{2}+y^{2}+z^{2}\leq r^{2}}{\text{ and }}z\geq 0\implies 0\leq z\leq {\sqrt {r^{2}-y^{2}-z^{2}}}$
Therefore, by generalized Fubini's theorem (triple integrals version) (i) and proposition about volume given by triple integration, the desired volume is ${\begin{aligned}\iiint _{D}1dV&=\int _{0}^{r}\int _{0}^{\sqrt {r^{2}-z^{2}}}\int _{0}^{\sqrt {r^{2}-y^{2}-z^{2}}}1\,dx\,dy\,dz\\&=\int _{0}^{r}\int _{0}^{\sqrt {r^{2}-z^{2}}}\left({\sqrt {r^{2}-y^{2}-z^{2}}}\right)\,dy\,dz\\&=\int _{0}^{r}\int _{0}^{\pi /2}\left(\left(\underbrace {\sqrt {r^{2}-z^{2}-\left(r^{2}-z^{2}\right)\sin ^{2}\theta }} _{{\sqrt {(r^{2}-z^{2})(1-\sin ^{2}\theta )}}={\sqrt {(r^{2}-z^{2})}}|\cos \theta |}\right)\left({\sqrt {r^{2}-z^{2}}}\right)\cos \theta \right)\,d\theta \,dz\qquad {\text{let }}y={\sqrt {r^{2}-z^{2}}}\sin \theta \implies dy=\left({\sqrt {r^{2}-z^{2}}}\right)\cos \theta \,d\theta {\text{ and }}\theta =\arcsin \left({\frac {y}{\sqrt {r^{2}-z^{2}}}}\right)({\text{by restricting the domain of sine function to }}-{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}})\\&=\int _{0}^{r}\left({\sqrt {r^{2}-z^{2}}}\right)^{2}\int _{0}^{\pi /2}(\cos \theta )(\cos \theta )\,d\theta \,dz\qquad |\cos \theta |=\cos \theta {\text{ because }}\cos \theta \geq 0{\text{ when }}{\frac {-\pi }{2}}\leq \theta \leq {\frac {\pi }{2}}\\&=\int _{0}^{r}\left(r^{2}-z^{2}\right)\int _{0}^{\pi /2}{\frac {1+\cos(2\theta )}{2}}\,d\theta \,dz\qquad {\text{by double angle formula}}\\&={\frac {1}{2}}\int _{0}^{r}\left(r^{2}-z^{2}\right)\left({\frac {\pi }{2}}+{\frac {1}{2}}\cdot \underbrace {\sin \pi } _{0}\right)\,dz\\&={\frac {\pi }{4}}\left[r^{2}z-{\frac {z^{3}}{3}}\right]_{0}^{r}\\&={\frac {\pi }{4}}\left(r^{3}-{\frac {r^{3}}{3}}\right)\\&={\frac {\pi }{4}}\cdot {\frac {2}{3}}\cdot r^{3}\\&={\frac {1}{6}}\pi r^{3}\end{aligned}}$
(ii) Because there are eight octants (in $\mathbb {R} ^{3}$), and each octant is symmetric to each other. There are seven more regions that are symmetric to $D$ in octants other than octant (+,+,+). Thus, the desired volume is $8\left({\frac {1}{6}}\pi r^{3}\right)={\frac {4}{3}}\pi r^{3}.$ $\Box$ Remark. The region mentioned in (ii) is a ball (solid figure bounded by a sphere) of radius $r$.
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