# Multiple Integrals

### Multiple Integrals:

1. Definition:

• Multiple integrals extend the concept of single integrals to functions of multiple variables. They involve integrating functions over regions in higher-dimensional spaces.
2. Types of Multiple Integrals:

a. Double Integrals:

• Involves integrating a function of two variables over a region in the plane.
• Denoted as ${\iint }_{R}f\left(x,y\right)\text{\hspace{0.17em}}dA$where $R$ is the region in the xy-plane 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 ${\iiint }_{V}f\left(x,y,z\right)\text{\hspace{0.17em}}dV$ 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.
3. 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 $f\left(x,y\right)$, the double integral ${\iint }_{R}f\left(x,y\right)\text{\hspace{0.17em}}dA$ can be expressed as ${\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)\text{\hspace{0.17em}}dy\text{\hspace{0.17em}}dx$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.
4. 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.

5. 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 $f\left(x,y\right)={x}^{2}+y$ 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 ${\iint }_{R}f\left(x,y\right)\text{\hspace{0.17em}}dA$.

1. Objective:

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

• Step 1: Setting Up the Integral:

• Express the double integral ${\iint }_{R}f\left(x,y\right)\text{\hspace{0.17em}}dA$ as ${\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)\text{\hspace{0.17em}}dy\text{\hspace{0.17em}}dx$ over the region $R$.
• For $f\left(x,y\right)={x}^{2}+y$, the integral becomes:
• Step 2: Evaluate the Inner Integral:

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

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

• The value of the double integral ${\iint }_{R}\left({x}^{2}+y\right)\text{\hspace{0.17em}}dy\text{\hspace{0.17em}}dx$ 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 $f\left(x,y,z\right)=x+y+z$ 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 ${\iiint }_{V}f\left(x,y,z\right)\text{\hspace{0.17em}}dV$.

1. Objective:

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

• Step 1: Setting Up the Integral:

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

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

• Integrate the result from the inner integral with respect to $y$ over the given interval: ${\int }_{0}^{2}\left(3x+3y+\frac{9}{2}\right)\text{\hspace{0.17em}}dy$ $={\left[3xy+\frac{3{y}^{2}}{2}+\frac{9y}{2}\right]}_{0}^{2}=\left(6x+6+9\right)-\left(0\right)$ $=6x+15$
• Step 4: Evaluate the Outer Integral:

• Integrate the result from the intermediate integral with respect to $x$ over the given interval: ${\int }_{0}^{1}\left(6x+15\right)\text{\hspace{0.17em}}dx$ $={\left[3{x}^{2}+15x\right]}_{0}^{1}=\left(3+15\right)-\left(0\right)$ = 18
3. Result:

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