Multiple Integrals: Double & Triple Integrals, Region Types, Order of Integration

1. Introduction to Multiple Integrals

Multiple integrals extend single-variable integration to functions of two or more variables. They are used to compute volumes under surfaces, mass of irregular objects, and probability over regions.

1.1 Double Integrals

A double integral evaluates a function $f(x,y)$ over a region $R$ in the $xy$-plane: $${\iint }_{R}f(x,y)dA=\underset{m,n\to \infty }{\lim }\sum _{i=1}^m\sum _{j=1}^{n}f(x_{ij}^{\ast },y_{ij}^{\ast })\Delta A_{ij}$$ Key Idea: Break the region into tiny rectangles, evaluate $f$ at each, and sum them up.

1.1.1 Types of Regions

Regions in double integrals are classified into Type I and Type II:

Worked Example 1: Determine the type of the region $R$ bounded by $y=x^2$, $y=4$, $x=-2$, and $x=2$. Solution:

• Since $y$ is expressed as a function of $x$, this is a Type I region.
• Bounds: $-2\le x\le 2$, $x^2\le y\le 4$.

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2. Setting Up Double Integrals

2.1 Iterated Integrals

Double integrals can be written as iterated integrals (repeated single integrals): $${\iint }_{R}f(x,y)dA={\int }_{x=a}^b\left({\int }_{y=g_1(x)}^{g_2(x)}f(x,y)dy\right)dx\text{(Type I)}$$ $$={\int }_{y=c}^d\left({\int }_{x=h_1(y)}^{h_2(y)}f(x,y)dx\right)dy\text{(Type II)}$$

Worked Example 2: Evaluate ${\iint }_R(x+y)dA$ where $R$ is the triangle with vertices $(0,0)$, $(1,0)$, and $(0,1)$. Solution:

• Type I: $0\le x\le 1$, $0\le y\le 1-x$
• Type II: $0\le y\le 1$, $0\le x\le 1-y$
• Compute using either order (both should give the same result): $${\int }_0^1{\int }_0^{1-x}(x+y)dydx={\int }_0^1{\left[xy+\frac{y^2}{2}\right]}_0^{1-x}dx={\int }_0^1\left(x(1-x)+\frac{(1-x{)}^2}{2}\right)dx$$ $$={\int }_0^1\left(x-x^2+\frac{1-2x+x^2}{2}\right)dx={\int }_0^1\left(\frac{x}{2}-\frac{x^2}{2}+\frac{1}{2}\right)dx=\frac{1}{4}$$

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3. Changing the Order of Integration

Sometimes, reversing the order of integration simplifies computation.

Worked Example 3: Evaluate ${\int }_0^1{\int }_{x^2}^x{e}^{y^2}dydx$ by reversing the order. Solution:

1. Sketch the region $R$:
   • $y=x^2$ (parabola)
   • $y=x$ (line)
   • $x=0$ to $x=1$
2. Reverse bounds:
   • For $y$ from $0$ to $1$, $x$ goes from $y$ to $\sqrt{y}$.
3. Rewrite the integral: $${\int }_0^1{\int }_y^{\sqrt{y}}{e}^{y^2}dxdy={\int }_0^1{e}^{y^2}(\sqrt{y}-y)dy$$
4. This may still require special functions, but the setup is now clearer.

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4. Triple Integrals

Triple integrals extend double integrals to three dimensions: $${\iiint }_{E}f(x,y,z)dV={\iiint }_{E}f(x,y,z)dxdydz$$ Bounds:

• Type I: $a\le x\le b$, $g_1(x)\le y\le g_2(x)$, $h_1(x,y)\le z\le h_2(x,y)$
• Type II: $c\le y\le d$, $h_1(y)\le x\le h_2(y)$, $k_1(x,y)\le z\le k_2(x,y)$
• Type III: $r_1\le r\le r_2$, $\alpha \le \theta \le \beta$, $z_1(r,\theta )\le z\le z_2(r,\theta )$ (cylindrical/spherical)

Worked Example 4: Find the volume of the solid bounded by $z=4-x^2-y^2$ and $z=0$. Solution:

• Project onto $xy$-plane: $x^2+y^2\le 4$ (circle of radius 2).
• Use cylindrical coordinates: $$V={\iiint }_{E}1dV={\int }_0^{2\pi }{\int }_0^2{\int }_0^{4-r^2}rdzdrd\theta$$ $$={\int }_0^{2\pi }{\int }_0^{2}r(4-r^2)drd\theta ={\int }_0^{2\pi }{\left[2r^2-\frac{r^4}{4}\right]}_0^{2}d\theta ={\int }_0^{2\pi }4d\theta =8\pi$$

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5. Applications of Multiple Integrals

5.1 Volume Under a Surface

For $z=f(x,y)$ over region $R$: $$V={\iint }_{R}f(x,y)dA$$

5.2 Mass of a Lamina

For density $\rho (x,y)$: $$M={\iint }_R\rho (x,y)dA$$

5.3 Average Value

$$f_{\text{avg}}=\frac{1}{A(R)}{\iint }_{R}f(x,y)dA$$

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## In the real world

1. eSewa (Nepal):

   • Idea Used: Double Integrals for Area Calculation
   • How: eSewa’s backend systems use spatial integration to calculate optimal delivery routes for couriers. The volume under a cost-surface (distance × fuel consumption) is minimized using double integrals to find the most efficient path, reducing delivery time and fuel costs.

2. Pathao (Nepal/Global):

   • Idea Used: Triple Integrals for Traffic Flow Modeling
   • How: Pathao’s algorithm for ride-matching uses 3D integration to model traffic density in real-time. By integrating speed, vehicle count, and time over a grid of city blocks, Pathao predicts congestion and suggests alternative routes. This is mathematically represented as: $${\iiint }_V\rho (x,y,t)dxdydt$$ where $\rho$ is traffic density at position $(x,y)$ and time $t$.

3. NTC (Nepal Telecom):

   • Idea Used: Double Integrals for Signal Coverage Optimization
   • How: NTC uses double integrals to model signal strength across a region. The integral of signal decay functions over a city map helps determine optimal tower placements. For example, the coverage area $A$ where signal $S(x,y)\ge \text{threshold}$ is computed as: $$A={\iint }_{R}H(S(x,y)-\text{threshold})dA$$ where $H$ is the Heaviside step function.

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## Visualizing Key Concepts

Figure 1: Type I vs. Type II Regions

Figure 2: Volume Under a Surface (Triple Integral)

Figure 3: Changing Order of Integration

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## Exam Tip

1. Always sketch the region before setting up bounds. Misidentifying Type I/II is a common mistake.
2. Check bounds carefully: A sign error in $g_1(x)$ or $g_2(x)$ can lead to incorrect results.
3. Practice reversing orders: Many problems are designed to test this skill. Look for symmetry or simpler bounds.
4. For triple integrals, decide early whether Cartesian, cylindrical, or spherical coordinates will simplify the problem.
5. Applications: Remember that volume, mass, and average value are frequent exam questions. Link integrals to physical quantities (e.g., "volume under a surface" = double integral of 1).
6. Jacobians: If using polar/cylindrical/spherical coordinates, include the Jacobian determinant (e.g., $r$ in polar, $r$ in cylindrical, ${\rho }^2\sin \varphi$ in spherical).

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Past Exam Question Solutions: a) Evaluate ${\int }_0^1{\int }_x^{1-x}xydydx$:

• Reverse order: $0\le y\le 1$, $0\le x\le 1-y$.
• Integral becomes ${\int }_0^1{\int }_0^{1-y}xydxdy$.
• Solve to get $\frac{1}{24}$.

b) Evaluate ${\iint }_R{e}^{x+y}dA$ where $R$ is the square $[0,1]\times [0,1]$:

• Separate variables: ${\int }_0^1{e}^{x}dx{\int }_0^1{e}^{y}dy=(e-1{)}^2$.

c) Volume of solid with base $R=\{(x,y)|x^2+y^2\le 1\}$ and height $z=x+y+2$:

• Use polar coordinates: $V={\int }_0^{2\pi }{\int }_0^1(r\cos \theta +r\sin \theta +2)rdrd\theta =3\pi$.