Multiple Integrals
Multiple Integrals:

Definition:
 Multiple integrals extend the concept of single integrals to functions of multiple variables. They involve integrating functions over regions in higherdimensional spaces.

Types of Multiple Integrals:
a. Double Integrals:
 Involves integrating a function of two variables over a region in the plane.
 Denoted as ${\iint}_{R}f(x,y)\text{\hspace{0.17em}}dA$where $R$ is the region in the xyplane and $dA$represents the differential area element.
 Computational methods include iterated integrals (Fubini's theorem) and polar, rectangular, or other coordinate transformations.
b. Triple Integrals:
 Extends the concept to integrate a function of three variables over a region in 3D space.
 Denoted as ${\iiint}_{V}f(x,y,z)\text{\hspace{0.17em}}dV$ where $V$ represents a volume in 3D space and $dV$is the differential volume element.
 Methods involve iterated integrals in various coordinate systems (rectangular, cylindrical, spherical) and applications in physics, engineering, and geometry.

Iterated Integrals:
a. Double Integrals as Iterated Integrals:
 Expresses a double integral as an iterated integral, computing one integral with respect to one variable and then the other.
 For a function $f(x,y)$, the double integral ${\iint}_{R}f(x,y)\text{\hspace{0.17em}}dA$ can be expressed as ${\int}_{a}^{b}{\int}_{c}^{d}f(x,y)\text{\hspace{0.17em}}dy\text{\hspace{0.17em}}dx$or vice versa.
b. Triple Integrals as Iterated Integrals:
 Similar to double integrals, expresses a triple integral as iterated integrals in various orders based on the application and convenience.

Applications:

Physics: Used in calculating mass, moments, center of mass, and physical quantities in 3D space (like electric charge distribution).

Engineering: Applied in solving problems related to fluid dynamics, heat transfer, and material science involving volumes and fluxes in three dimensions.

Probability and Statistics: Utilized to calculate joint probabilities, expected values, and moments in multiple dimensions.


Properties of Multiple Integrals:
 Linearity, additivity, and properties of single integrals extend to multiple integrals.
 Changing the order of integration often simplifies calculations, especially when dealing with certain regions or coordinate systems.
Example: Double Integral Calculation
Consider the function $f(x,y)={x}^{2}+y$ over the region $R$ defined by $0\le x\le 2$and $1\le y\le 3$. Let's find the value of the double integral ${\iint}_{R}f(x,y)\text{\hspace{0.17em}}dA$.

Objective:
 Calculate the double integral of $f(x,y)={x}^{2}+y$ over the region $R$ defined by $0\le x\le 2$ and $1\le y\le 3$.

Solution Steps:

Step 1: Setting Up the Integral:
 Express the double integral ${\iint}_{R}f(x,y)\text{\hspace{0.17em}}dA$ as ${\int}_{a}^{b}{\int}_{c}^{d}f(x,y)\text{\hspace{0.17em}}dy\text{\hspace{0.17em}}dx$ over the region $R$.
 For $f(x,y)={x}^{2}+y$, the integral becomes: ${\iint}_{R}({x}^{2}+y)\text{\hspace{0.17em}}dy\text{\hspace{0.17em}}dx\text{over}0\le x\le 2\text{and}1\le y\le 3$

Step 2: Evaluate the Inner Integral:
 Integrate with respect to $y$ first: ${\int}_{1}^{3}({x}^{2}+y)\text{\hspace{0.17em}}dy$ $={[{x}^{2}y+\frac{{y}^{2}}{2}]}_{1}^{3}=(3{x}^{2}+\frac{9}{2})({x}^{2}+\frac{1}{2})$ $=2{x}^{2}+4$

Step 3: Evaluate the Outer Integral:
 Integrate the result from the inner integral with respect to $x$ over the given interval: ${\int}_{0}^{2}(2{x}^{2}+4)\text{\hspace{0.17em}}dx$ $={[\frac{2{x}^{3}}{3}+4x]}_{0}^{2}=(\frac{16}{3}+8)(0)$ $=\frac{40}{3}$


Result:
 The value of the double integral ${\iint}_{R}({x}^{2}+y)\text{\hspace{0.17em}}dy\text{\hspace{0.17em}}dx$ over the region $R$ defined by $0\le x\le 2$ and $1\le y\le 3$ is $\frac{40}{3}$.
Example: Triple Integral Calculation
Consider the function $f(x,y,z)=x+y+z$ over the region $V$ defined by $0\le x\le 1$, $0\le y\le 2$, and $0\le z\le 3$. Let's find the value of the triple integral ${\iiint}_{V}f(x,y,z)\text{\hspace{0.17em}}dV$.

Objective:
 Calculate the triple integral of $f(x,y,z)=x+y+z$ over the region $V$ defined by $0\le x\le 1$, $0\le y\le 2$, and $0\le z\le 3$.

Solution Steps:

Step 1: Setting Up the Integral:
 Express the triple integral ${\iiint}_{V}f(x,y,z)\text{\hspace{0.17em}}dV$ as ${\int}_{a}^{b}{\int}_{c}^{d}{\int}_{p}^{q}f(x,y,z)\text{\hspace{0.17em}}dz\text{\hspace{0.17em}}dy\text{\hspace{0.17em}}dx$ over the region $V$.
 For $f(x,y,z)=x+y+z$ and the given limits, the integral becomes: ${\iiint}_{V}(x+y+z)\text{\hspace{0.17em}}dz\text{\hspace{0.17em}}dy\text{\hspace{0.17em}}dx\text{over}0\le x\le 1,\text{\hspace{0.17em}}0\le y\le 2,\text{\hspace{0.17em}}0\le z\le 3$

Step 2: Evaluate the Inner Integral:
 Integrate with respect to $z$ first: ${\int}_{0}^{3}(x+y+z)\text{\hspace{0.17em}}dz$ $={[xz+yz+\frac{{z}^{2}}{2}]}_{0}^{3}=(3x+3y+\frac{9}{2})(0)$ $=3x+3y+\frac{9}{2}$

Step 3: Evaluate the Intermediate Integral:
 Integrate the result from the inner integral with respect to $y$ over the given interval: ${\int}_{0}^{2}(3x+3y+\frac{9}{2})\text{\hspace{0.17em}}dy$ $={[3xy+\frac{3{y}^{2}}{2}+\frac{9y}{2}]}_{0}^{2}=(6x+6+9)(0)$ $=6x+15$

Step 4: Evaluate the Outer Integral:
 Integrate the result from the intermediate integral with respect to $x$ over the given interval: ${\int}_{0}^{1}(6x+15)\text{\hspace{0.17em}}dx$ $={[3{x}^{2}+15x]}_{0}^{1}=(3+15)(0)$


Result:
 The value of the triple integral ${\iiint}_{V}(x+y+z)\text{\hspace{0.17em}}dz\text{\hspace{0.17em}}dy\text{\hspace{0.17em}}dx$over the region $V$ defined by $0\le x\le 1$, $0\le y\le 2$, and $0\le z\le 3$ is $18$.