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Double Integrals
Definition If we have a well behaved function ^[1] $f(x,y)$ in a region $R$ then the double integral in that region is defined as -- $$\int\int_{R} f(x,y)\ dA$$ Where $dA$ is the elementary area in the region $R$. Generally, double integral of the function $f(x,y)$ is beautifully defined as the limit of Riemann sums^[2] in the region $R$ -- $$\lim_{n \to \infty}\sum_{i=1}^{n} f(x_i,y_i)\ \Delta A_i$$ Where, the region $R$ is divided into n small regions , i-th element having area $\Delta A_i$ and in that element at some point $(x_i,y_i)$ the function has the value $f(x_i,y_i)$. -1
Motivation We have been using integration for finding area, mass, the moment of inertia, and so on. The use case of integration is indispensable. And as we jump into the realms of functions with more than one variable, it is natural to ponder on how integration - as a process - will evolve. Double Integral is the first step towards Multiple Integral, which has immense applications in probability theory, Turbulent Flows, improvement of combustion engines, and more in engineering and architecture ^[3]. In single variable calculus, typically we thought of integration as the area under the curve - similarly, building up to the same analogy, we'll see how the double integral can be interpreted as the volume under a surface. -1
Bird's Eye View A single variable function $f(x)$ can be represented as a curve in X-Y plane. For $f(x)$ the definite integral $\int_{a}^{b} f(x) dx$ represents area under the curve between $a\leq x \leq b$ . -1
-1
Animation 1 : $\int f(x) dx$ is the area under the curve $y=f(x)$ and it can be represented as Riemann sum of infinitesimally thin rectangular areas.
For function $f(x,y)$ the double integral $\int_{a}^{b}\int_{c}^{d} f(x,y)\ dx dy$ represent a volume under the surface in the region $R$ bounded by limits $a\leq x \leq b$ and $c\leq y \leq d$. 0
0
Animation 2: $\iint f(x,y)\,dxdy$ is the volume under the surface $z=f(x,y)$ and it can be approximated as sum of volume of cubiods.
Usually, we use the integration formulas ^[5] to integrate a function but sometimes we encounter some stubborn functions which we cannot integrate by any method using the formulas. There comes the Riemann's sum method which is applicable in any situation, and it is based on calculating the area or volume by dividing the region into infinitesimal elements, but this is a numerical approach - we need computational power and it is quite straightforward. For doing double integrals by non-numerical methods with help of anti-derivative formulas^[5](i.e.$\int x^n dx= x^{n+1}/(n+1)$ ) we first set one of the variables (Let $y$) as a spectator and do the integration over the other variable (here $x$). $$\int_c^d\left[\int_a^b f(x,y)dx\right]dy = \int_c^d\left[g(x,y){\large \|}_a^b\right]dy= \int_c^dg(y)dy$$ It serves as a single integral and gives us a "thick" area as a function of $y$ (here $g(y)\times dy$, the blue plane in the below animation), with some infinitesimal thickness $dy$. Then we do the integration over $y$ as if we are collecting that "thick" areas along $y$, to get the whole volume under the surface of the function in the integration region. This is called the iteration method. 0
0
Animation 3 : Double integration process of the function $z=f(x,y)$ in the region $R$
Context of the Definition To get the taste of it let's start with a function $f(x,y)=1+x+y $ to be integrated in the limit $0\leq x\leq 1 , 0\leq y\leq 2$. For the sake of seeing it in our coordinate system, we assign $z=f(x,y)$. To make it more meaningful (remembering the Riemann Sum), we restate the problem as - Find the volume of the region bounded by the planes $z=1+x+y,\ y=2,\ x=1$ and coordinate planes. Setting up the double integral here will be $$\int_{x=0}^{1}\int_{y=0}^{2}(1+x+y)\ dydx$$ Let's first integrate over $y$ treating $x$ as some constant (process called iterated integration) -- $$\int_{x=0}^{1}\left(y+xy+\dfrac{y^2}{2}\right)\Bigg\|_{0}^{2}\ dx = \int_{x=0}^{1}(4+2x)\ dx$$ The function $(4+2x)$ gives the area under the surface at some arbitrary $x$, and multiplied by $dx$ it gives a thin "slab" like volume! Now if we sum over all such "slabs" we will get the total volume; to do it we just have to integrate over $x$ . $$\int_{0}^{1}(4+2x)\ dx = (4x+x^2)\bigg\|_{0}^{1} =5 $$ This is the value of the double integral and it is also the volume of the region in the restated problem. Thus we get the recipe and also the physical interpretation of evaluating a Double Integral. The same recipe is valid for higher dimensional integral but there the physical meaning becomes cumbersome. 0
0
Animation 4: Integration process of $z=(1+x+y)$ in the previous example.
♦ Properties of the Double Integral $$\int\int_{R}(f(x,y)+g(x,y))\ dA=\int\int_{R}f(x,y)\ dA+\int\int_{R}g(x,y)\ dA$$ If $\alpha$ is a constant, $$\int\int_{R}\alpha\ f(x,y)\ dA=\alpha \int\int_{R}f(x,y)\ dA$$. The integral can be done by splitting the region into multiple regions. Here region $R$ is split into two regions $R1$ and $R2$, $$\int\int_{R}f(x,y)\ dA=\int\int_{R2}f(x,y)\ dA+\int\int_{R1}f(x,y)\ dA$$ . ♦ Integration in non-rectangular regions The recipe given before is incomplete because we do not always have so nice constant values as limits, creating a rectangular region. Setting the limits for non-rectangular regions can be quite tricky. In those cases, it is best to draw the region and see what curves bound the region and serve as limits of integration. Lets see it with an example - again we choose the function $f(x,y)=(1+x+y)$ but now we have to get the volume of the region bounded by $z=f(x,y)$, $2x+y=2$ plane and coordinate planes . At first, we integrate over $y$, but we have to be cautious about the limit, as at different $x$ values the limit of $y$ is not a constant rather it is a function of $x$ $\rightarrow y=2-2x$. So here the integral becomes-- $$\int_{x=0}^{1}\int_{y=0}^{2-2x}(1+x+y)\ dydx$$ 0
0
The limit of $y$ is dependent on $x$
Evaluating the integral- \begin{align} \int_{x=0}^{1}\left(y+xy+\dfrac{y^2}{2}\right)\Bigg\|_{0}^{2-2x}\ dx & = \int_{0}^{1}\left((2-2x)+x(2-2x)+\dfrac{(2-2x)^2}{2}\right) dx \\ &= \int_{0}^{1}4(1-x)\ dx\\ & = 2\end{align} Now if we have wanted to do the $x$ integral first then the setup would have been like- $$\int_{y=0}^{2}\int_{x=0}^{1-\frac{y}{2}} (1+x+y)\ dxdy$$ and one can verify that it also gives the same answer. Generally, the limits can be any function rather than being a straight line(For more see Ref:[6]). In those cases, it is best to draw the region and choose whether the $x$ integration or the $y$ integration should be done first! Choosing the integration order wisely can reduce the amount of work significantly and occasionally it happens that the function can't be integrated in the other way. Property 3. is also a useful tool. In some cases, it may happen that the boundary can't be expressed as a single function but we can divide the region to get some good functions. In real-life applications we have to approximate a region with some functions -- functions are not ready-made. 0
0
Integrating over non-rectengular region and division of the region to get some meaningful functions
♦ Using the Polar Coordinate System So far we used Cartesian Coordinate systems ( X-Y axis perpendicular on each other) but we can use any coordinate systems as feasible to the given problem. One of such most used coordinate system is polar coordinates. Probably you have noticed, "$dA$" used in the definition was substituted by "$dy\ dx$" in the previous examples. It is that because in Cartesian coordinate systems elementary area is a rectangle with sides "$dx$" and "$dy$". Similarly, we can find that in polar coordinates the elementary area is "$r\ dr\ d\theta$". 0
0
Elementary area expressed in rectangular and polar coordinates in a region.
Let's see with an example how polar coordinates can be used for integration-- We again choose $f(x,y)=1+x+y$ and now we have to find the volume of the region bounded by $z=f(x,y)$ , $x^2+y^2=1$ and coordinate planes. In the spirit of previous sections the integral should be-- $$\int_{x=-1}^{1}\int_{y=-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(1+x+y)\ dydx$$ Although we can do the integral this looks so messy. But we see that the region is circular as $x^2+y^2=1$ is an equation of a circle so we can try polar coordinates. The transformation between polar^[7] and cartesian coordinate is $$ x=r\cos\theta ,\quad y=r\sin\theta$$ So we transform everything to polar coordinates, $f(x,y)\rightarrow f(r,\theta)=1+r(\cos\theta+\sin\theta)$, $dxdy\rightarrow r\ dr\ d\theta$. For setting the limits it is better to draw/imagine the region. 0
So in polar coordinates the integral becomes-- $$\int_{r=0}^{1}\int_{\theta=0}^{2\pi}(1+r\cos\theta+r\sin\theta)\ rdrd\theta$$ Observe that here the limits are independent of each other. \begin{align} \int_{0}^{1}rdr\int_{0}^{2\pi}d\theta+\int_{0}^{1}r^2dr\int_{0}^{2\pi}\cos\theta\ d\theta+\int_{0}^{1}r^2dr\int_{0}^{2\pi}\sin\theta\ d\theta =\pi +0+0=\pi\end{align} You can cross-check the answer by doing the integral set with $x, y$. So, if you find some rotational symmetry try to use polar coordinates to evaluate the integral; it is not guaranteed that this will simplify the problem in every situation. Applications We have already seen that using double integral we can calculate the volume of a region. Next we calculate center of mass($\overline x,\overline y$) of a laminar object with mass density $\rho(x,y)=1+y$ and sides $y=0, y=x\ [0\leq x\leq2], y=(6-2x)\ [2\leq x \leq 3]$. From the definition of center of mass, $\overline x=\dfrac{1}{M}\int x dm, \overline y=\dfrac{1}{M}\int y dm$, Where $M=\int dm$ is the total mass of the lamina. \begin{align}M=\int dm &=\int\int\rho dA\\ &=\int_{0}^{2}\int_{0}^{x}(1+y)dydx+\int_{2}^{3}\int_{0}^{6-2x}(1+y)dydx\\ &=\int_{0}^{2}(x+\dfrac{x^2}{2})dx+\int_{2}^{3}((6-2x)+\dfrac{(6-2x)^2}{2})dx\\ &=5\end{align} You can see that property 1 & 3 is being used for evaluating the integrals. \begin{align}\overline x &=\dfrac{1}{M}\int_{x=0}^{3}\int_{y=0}^{f(x)}x(1+y) dydx\\ &=\dfrac{1}{M}\left[\int_{0}^{2}\int_{0}^{x}x(1+y) dydx+\int_{2}^{3}\int_{0}^{6-2x}x(1+y) dydx\right] \\ &=\dfrac{1}{5}\left(\dfrac{14}{3}+\dfrac{23}{6}\right) \\ &=1.7\end{align} similarly, \begin{align}\overline y &=\dfrac{1}{M}\left[\int_{0}^{2}\int_{0}^{x}y(1+y) dydx+\int_{2}^{3}\int_{0}^{6-2x}y(1+y) dydx\right] \\ &=\dfrac{1}{5}\left(\dfrac{8}{3}+\dfrac{4}{3}\right)\\ &=0.8\end{align} 0
0
The center of mass of the laminar object in the example
Do the integrals to get the answers by yourself? So we get the center of mass of the given lamina at $(\bar x=1.7,\bar y=0.8)$. It is an averaging process where elementary masses are multiplied by its distance from the axes. In a similar fashion, we can do statistical and probability calculations.(See Ref:[3] ,page-362 and Ref:[8]) In the calculation of surface integrals, double integral is the tool. 0
History Throughout the 17th and 18th centuries Mathematicians were coming closer to the idea of Riemannian Integration and in the 19th century after the discovery of limits Cauchy first tried to make integral calculus more meaningful. Next in 1854 Riemann Integration came from Riemann's paper on trigonometric series. Gaston Darboux made the idea more clear & modified by defining the convergence of the Upper Riemann and Lower Riemann sums but then Dirichlet showed that not all functions are Riemann integrable. In this situation, Lebesgue introduced an alternative form of integration in 1902. Lebesgue's summarized his idea in a letter to Paul Montel: I have to pay a certain sum, which I have collected in my pocket. I take the bills and coins out of my pocket and give them to the creditor in the order I find them until I have reached the total sum. This is the Riemann integral. But I can proceed differently. After I have taken all the money out of my pocket I order the bills and coins according to identical values and then I pay the several heaps one after the other to the creditor. This is my integral. - (Source: Ref :[9]) Lebesgue's work is completely abstract and showed that Mathematics has marched far from merely being a tool for science. 0
Pause and Ponder Think that integration is nothing but the sum of the values of the function at different points multiplied with the small areas/line-segments where the points reside. To do that by hand we have a shortcut (powered by calculus) --the antiderivative formulas. How can we think about the geometric interpretation of double integrals in Cylindrical Coordinates (or any other non-euclidean coordinate system)? 0
References https://ocw.mit.edu/ans7870/18/18.013a/textbook/HTML/chapter20/section04.html (non_integrable functions) Weisstein, Eric W. "Riemann Sum." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/RiemannSum.html Hubbard, John Hamal,Barbara Burke. “Integration.” Vector Calculus, Linear Algebra, And Differential Forms -A Unified Approach., Prentice Hall https://openstax.org/books/calculus-volume-3/pages/5-1-double-integrals-over-rectangular-regions http://www.pas.rochester.edu/~arijit/c02.pdf http://tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx Stover, Christopher and Weisstein, Eric W. "Polar Coordinates." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/PolarCoordinates.html https://web.ma.utexas.edu/users/m408m/Display15-5-7.shtml Siegmund-Schultze, Reinhard (2008), "Henri Lebesgue", in Timothy Gowers, June Barrow-Green, Imre Leader (eds.), Princeton Companion to Mathematics, Princeton University Press. http://tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-integrals-and-line-integrals-in-the-plane/part-a-double-integrals http://tutorial.math.lamar.edu/Classes/CalcI/AreaProblem.aspx 0
Further Reading For historical accounts, real-life examples and interesting problems: Howard Anton, Irl Bivens, Stephan Davis. CALCULUS Early Transcendentals, 10th Ed., John Wiley & Sons. Hubbard, John Hamal, Barbara Burke. “Integration.” Vector Calculus, Linear Algebra, And Differential Forms -A Unified Approach., Prentice Hall Tom M. Apostal, Calculus-I and Calculus-II, John Wiley & Sons, New Delhi. James Stewart, Multivariable Calculus, Concepts and Contexts, 2nd Ed., Brooks /Cole, Thomson Learning, USA, 2001. Surface Integrals Numerical Integration methods: Simpson's Rule, Product Rule, Monte-Carlo Method etc. Difficulties of Riemann sum Method in higher dimensions. 0

Double Integrals

Definition

If we have a well behaved function ^[1] $f(x,y)$ in a region $R$ then the double integral in that region is defined as --
$$\int\int_{R} f(x,y)\ dA$$
Where $dA$ is the elementary area in the region $R$.

Generally, double integral of the function $f(x,y)$ is beautifully defined as the limit of Riemann sums^[2] in the region $R$ -- $$\lim_{n \to \infty}\sum_{i=1}^{n} f(x_i,y_i)\ \Delta A_i$$ Where, the region $R$ is divided into n small regions , i-th element having area $\Delta A_i$ and in that element at some point $(x_i,y_i)$ the function has the value $f(x_i,y_i)$.

-1

Motivation

We have been using integration for finding area, mass, the moment of inertia, and so on. The use case of integration is indispensable. And as we jump into the realms of functions with more than one variable, it is natural to ponder on how integration - as a process - will evolve.

Double Integral is the first step towards Multiple Integral, which has immense applications in probability theory, Turbulent Flows, improvement of combustion engines, and more in engineering and architecture ^[3]. In single variable calculus, typically we thought of integration as the area under the curve - similarly, building up to the same analogy, we'll see how the double integral can be interpreted as the volume under a surface.

-1

Bird's Eye View

A single variable function $f(x)$ can be represented as a curve in X-Y plane. For $f(x)$ the definite integral $\int_{a}^{b} f(x) dx$ represents area under the curve between $a\leq x \leq b$ .

-1

Animation 1 : $\int f(x) dx$ is the area under the curve $y=f(x)$ and it can be represented as Riemann sum of infinitesimally thin rectangular areas.

For function $f(x,y)$ the double integral $\int_{a}^{b}\int_{c}^{d} f(x,y)\ dx dy$ represent a volume under the surface in the region $R$ bounded by limits $a\leq x \leq b$ and $c\leq y \leq d$.

Animation 2: $\iint f(x,y)\,dxdy$ is the volume under the surface $z=f(x,y)$ and it can be approximated as sum of volume of cubiods.

Usually, we use the integration formulas ^[5] to integrate a function but sometimes we encounter some stubborn functions which we cannot integrate by any method using the formulas. There comes the Riemann's sum method which is applicable in any situation, and it is based on calculating the area or volume by dividing the region into infinitesimal elements, but this is a numerical approach - we need computational power and it is quite straightforward.

For doing double integrals by non-numerical methods with help of anti-derivative formulas^[5](i.e.$\int x^n dx= x^{n+1}/(n+1)$ ) we first set one of the variables (Let $y$) as a spectator and do the integration over the other variable (here $x$).
$$\int_c^d\left[\int_a^b f(x,y)dx\right]dy = \int_c^d\left[g(x,y){\large |}_a^b\right]dy= \int_c^dg(y)dy$$
It serves as a single integral and gives us a "thick" area as a function of $y$ (here $g(y)\times dy$, the blue plane in the below animation), with some infinitesimal thickness $dy$.

Then we do the integration over $y$ as if we are collecting that "thick" areas along $y$, to get the whole volume under the surface of the function in the integration region. This is called the iteration method.

Animation 3 : Double integration process of the function $z=f(x,y)$ in the region $R$

Context of the Definition

To get the taste of it let's start with a function $f(x,y)=1+x+y $ to be integrated in the limit $0\leq x\leq 1 , 0\leq y\leq 2$. For the sake of seeing it in our coordinate system, we assign $z=f(x,y)$. To make it more meaningful (remembering the Riemann Sum), we restate the problem as - Find the volume of the region bounded by the planes $z=1+x+y,\ y=2,\ x=1$ and coordinate planes.
Setting up the double integral here will be

$$\int_{x=0}^{1}\int_{y=0}^{2}(1+x+y)\ dydx$$

Let's first integrate over $y$ treating $x$ as some constant (process called iterated integration) --

$$\int_{x=0}^{1}\left(y+xy+\dfrac{y^2}{2}\right)\Bigg|_{0}^{2}\ dx = \int_{x=0}^{1}(4+2x)\ dx$$

The function $(4+2x)$ gives the area under the surface at some arbitrary $x$, and multiplied by $dx$ it gives a thin "slab" like volume! Now if we sum over all such "slabs" we will get the total volume; to do it we just have to integrate over $x$ .

$$\int_{0}^{1}(4+2x)\ dx = (4x+x^2)\bigg|_{0}^{1} =5 $$
This is the value of the double integral and it is also the volume of the region in the restated problem. Thus we get the recipe and also the physical interpretation of evaluating a Double Integral. The same recipe is valid for higher dimensional integral but there the physical meaning becomes cumbersome.

Animation 4: Integration process of $z=(1+x+y)$ in the previous example.

♦ Properties of the Double Integral

$$\int\int_{R}(f(x,y)+g(x,y))\ dA=\int\int_{R}f(x,y)\ dA+\int\int_{R}g(x,y)\ dA$$
If $\alpha$ is a constant, $$\int\int_{R}\alpha\ f(x,y)\ dA=\alpha \int\int_{R}f(x,y)\ dA$$.
The integral can be done by splitting the region into multiple regions. Here region $R$ is split into two regions $R1$ and $R2$, $$\int\int_{R}f(x,y)\ dA=\int\int_{R2}f(x,y)\ dA+\int\int_{R1}f(x,y)\ dA$$ .

♦ Integration in non-rectangular regions

The recipe given before is incomplete because we do not always have so nice constant values as limits, creating a rectangular region. Setting the limits for non-rectangular regions can be quite tricky. In those cases, it is best to draw the region and see what curves bound the region and serve as limits of integration. Lets see it with an example - again we choose the function $f(x,y)=(1+x+y)$ but now we have to get the volume of the region bounded by $z=f(x,y)$, $2x+y=2$ plane and coordinate planes . At first, we integrate over $y$, but we have to be cautious about the limit, as at different $x$ values the limit of $y$ is not a constant rather it is a function of $x$ $\rightarrow y=2-2x$. So here the integral becomes--
$$\int_{x=0}^{1}\int_{y=0}^{2-2x}(1+x+y)\ dydx$$

The limit of $y$ is dependent on $x$

Evaluating the integral-
\begin{align} \int_{x=0}^{1}\left(y+xy+\dfrac{y^2}{2}\right)\Bigg|_{0}^{2-2x}\ dx & = \int_{0}^{1}\left((2-2x)+x(2-2x)+\dfrac{(2-2x)^2}{2}\right) dx \\
&= \int_{0}^{1}4(1-x)\ dx\\ & = 2\end{align}

Now if we have wanted to do the $x$ integral first then the setup would have been like-
$$\int_{y=0}^{2}\int_{x=0}^{1-\frac{y}{2}} (1+x+y)\ dxdy$$

and one can verify that it also gives the same answer.

Generally, the limits can be any function rather than being a straight line(For more see Ref:[6]). In those cases, it is best to draw the region and choose whether the $x$ integration or the $y$ integration should be done first! Choosing the integration order wisely can reduce the amount of work significantly and occasionally it happens that the function can't be integrated in the other way.

Property 3. is also a useful tool. In some cases, it may happen that the boundary can't be expressed as a single function but we can divide the region to get some good functions. In real-life applications we have to approximate a region with some functions -- functions are not ready-made.

Integrating over non-rectengular region and division of the region to get some meaningful functions

♦ Using the Polar Coordinate System

So far we used Cartesian Coordinate systems ( X-Y axis perpendicular on each other) but we can use any coordinate systems as feasible to the given problem. One of such most used coordinate system is polar coordinates. Probably you have noticed, "$dA$" used in the definition was substituted by "$dy\ dx$" in the previous examples. It is that because in Cartesian coordinate systems elementary area is a rectangle with sides "$dx$" and "$dy$". Similarly, we can find that in polar coordinates the elementary area is "$r\ dr\ d\theta$".

Elementary area expressed in rectangular and polar coordinates in a region.

Let's see with an example how polar coordinates can be used for integration--

We again choose $f(x,y)=1+x+y$ and now we have to find the volume of the region bounded by $z=f(x,y)$ , $x^2+y^2=1$ and coordinate planes. In the spirit of previous sections the integral should be--
$$\int_{x=-1}^{1}\int_{y=-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(1+x+y)\ dydx$$
Although we can do the integral this looks so messy. But we see that the region is circular as $x^2+y^2=1$ is an equation of a circle so we can try polar coordinates. The transformation between polar^[7] and cartesian coordinate is $$ x=r\cos\theta ,\quad y=r\sin\theta$$

So we transform everything to polar coordinates, $f(x,y)\rightarrow f(r,\theta)=1+r(\cos\theta+\sin\theta)$, $dxdy\rightarrow r\ dr\ d\theta$. For setting the limits it is better to draw/imagine the region.

So in polar coordinates the integral becomes--
$$\int_{r=0}^{1}\int_{\theta=0}^{2\pi}(1+r\cos\theta+r\sin\theta)\ rdrd\theta$$
Observe that here the limits are independent of each other.
\begin{align} \int_{0}^{1}rdr\int_{0}^{2\pi}d\theta+\int_{0}^{1}r^2dr\int_{0}^{2\pi}\cos\theta\ d\theta+\int_{0}^{1}r^2dr\int_{0}^{2\pi}\sin\theta\ d\theta =\pi +0+0=\pi\end{align}
You can cross-check the answer by doing the integral set with $x, y$.

So, if you find some rotational symmetry try to use polar coordinates to evaluate the integral; it is not guaranteed that this will simplify the problem in every situation.

Applications

We have already seen that using double integral we can calculate the volume of a region. Next we calculate center of mass($\overline x,\overline y$) of a laminar object with mass density $\rho(x,y)=1+y$ and sides $y=0, y=x\ [0\leq x\leq2], y=(6-2x)\ [2\leq x \leq 3]$.
From the definition of center of mass, $\overline x=\dfrac{1}{M}\int x dm, \overline y=\dfrac{1}{M}\int y dm$, Where $M=\int dm$ is the total mass of the lamina.

\begin{align}M=\int dm &=\int\int\rho dA\\ &=\int_{0}^{2}\int_{0}^{x}(1+y)dydx+\int_{2}^{3}\int_{0}^{6-2x}(1+y)dydx\\ &=\int_{0}^{2}(x+\dfrac{x^2}{2})dx+\int_{2}^{3}((6-2x)+\dfrac{(6-2x)^2}{2})dx\\ &=5\end{align}
You can see that property 1 & 3 is being used for evaluating the integrals.

\begin{align}\overline x &=\dfrac{1}{M}\int_{x=0}^{3}\int_{y=0}^{f(x)}x(1+y) dydx\\ &=\dfrac{1}{M}\left[\int_{0}^{2}\int_{0}^{x}x(1+y) dydx+\int_{2}^{3}\int_{0}^{6-2x}x(1+y) dydx\right] \\ &=\dfrac{1}{5}\left(\dfrac{14}{3}+\dfrac{23}{6}\right) \\ &=1.7\end{align}
similarly,
\begin{align}\overline y &=\dfrac{1}{M}\left[\int_{0}^{2}\int_{0}^{x}y(1+y) dydx+\int_{2}^{3}\int_{0}^{6-2x}y(1+y) dydx\right] \\ &=\dfrac{1}{5}\left(\dfrac{8}{3}+\dfrac{4}{3}\right)\\ &=0.8\end{align}

The center of mass of the laminar object in the example

Do the integrals to get the answers by yourself? So we get the center of mass of the given lamina at $(\bar x=1.7,\bar y=0.8)$.

It is an averaging process where elementary masses are multiplied by its distance from the axes. In a similar fashion, we can do statistical and probability calculations.(See Ref:[3] ,page-362 and Ref:[8])

In the calculation of surface integrals, double integral is the tool.

History

Throughout the 17th and 18th centuries Mathematicians were coming closer to the idea of Riemannian Integration and in the 19th century after the discovery of limits Cauchy first tried to make integral calculus more meaningful. Next in 1854 Riemann Integration came from Riemann's paper on trigonometric series. Gaston Darboux made the idea more clear & modified by defining the convergence of the Upper Riemann and Lower Riemann sums but then Dirichlet showed that not all functions are Riemann integrable. In this situation, Lebesgue introduced an alternative form of integration in 1902. Lebesgue's summarized his idea in a letter to Paul Montel:

I have to pay a certain sum, which I have collected in my pocket. I take the bills and coins out of my pocket and give them to the creditor in the order I find them until I have reached the total sum. This is the Riemann integral. But I can proceed differently. After I have taken all the money out of my pocket I order the bills and coins according to identical values and then I pay the several heaps one after the other to the creditor. This is my integral. - (Source: Ref :[9])

Lebesgue's work is completely abstract and showed that Mathematics has marched far from merely being a tool for science.

Pause and Ponder

Think that integration is nothing but the sum of the values of the function at different points multiplied with the small areas/line-segments where the points reside. To do that by hand we have a shortcut (powered by calculus) --the antiderivative formulas.

How can we think about the geometric interpretation of double integrals in Cylindrical Coordinates (or any other non-euclidean coordinate system)?

References

https://ocw.mit.edu/ans7870/18/18.013a/textbook/HTML/chapter20/section04.html (non_integrable functions)
Weisstein, Eric W. "Riemann Sum." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/RiemannSum.html
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