Multivariable Calculus Study Notes

Triple Integrals

• Triple integrals extend the concept of double integrals to three dimensions.

Triple Integral Setup

• Partitioning: Divide the x, y, and z axes into partitions, creating mini boxes.
• Sample Point: Select a sample point within each mini box.
• Evaluation: Evaluate the function $f$ at the sample point.
• Volume Element: Multiply the function value by the volume element $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\Delta}x{\Delta}y{\Delta}z)$.
• Riemann Sum: Sum these products over all mini boxes.
• The result is a four-dimensional quantity (function value + three dimensions).

Integral Notation

• Triple integral over a 3D region D: $\int\int\int_{D} dV$
• Iterated integral (Fubini-like version):
  $\int<em>{A}^{B} \int</em>{C}^{D} \int_{R}^{S} f(x, y, z) dz dy dx$

Review question

Compute the double integral $\iint_{R} \arctan(y/x) dA$
where R is defined by:

• $1 \leq x^2 + y^2 \leq 4$ (concentric circles with radii 1 and 2)
• $0 \leq y \leq x$

Solution Steps

• Polar Coordinates: Convert to polar coordinates:

  • $x = r \cos(\theta)$
  • $y = r \sin(\theta)$
  • $dA = r dr d\theta$

• Region in Polar Coordinates:

  • $1 \leq r \leq 2$
  • $0 \leq \theta \leq \frac{\pi}{4}$

Integral Setup

• Original integral: $\iint_{R} \arctan(\frac{y}{x}) dA$

• Substitution: $\arctan(\frac{r\sin(\theta)}{r\cos(\theta)}) = \arctan(\tan(\theta)) = \theta$

• Transformed integral: $\int<em>{0}^{\frac{\pi}{4}} \int</em>{1}^{2} \theta \cdot r dr d\theta$

Separable Integrals

$\int<em>{1}^{2} r dr \cdot \int</em>{0}^{\frac{\pi}{4}} \theta d\theta$

• Integrating with respect to r:

$\int<em>{1}^{2} r dr = \frac{r^2}{2} \bigg|</em>{1}^{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$

Integrating with respect to $\theta$:
$\int<em>{0}^{\frac{\pi}{4}} \theta d\theta = \frac{\theta^2}{2} \bigg|</em>{0}^{\frac{\pi}{4}} = \frac{(\frac{\pi}{4})^2}{2} = \frac{\pi^2}{32}$

• Final Result:
  $(\frac{3}{2}) (\frac{\pi^2}{32}) = \frac{3\pi^2}{64}$

Generalized Rectangles

Review of double integrals over a generalized rectangular region.

Problem Statement

Compute $\iint_{D} y^2 dA$

where D is a triangle with vertices (0,1), (1,2), and (4,1).

Type I Region

Can divide the triangular region into two regions using the altitude from vertex (1,2) to the base.
However it will lead into two regions. From x=0 to x=1, and from x=1 tp x=4.

For type one, the equations are:

Lower curve: y=1
Upper curve: y = x + 1
With bounds: $0 \leq x \leq 1$

Additional setup:

Upper curve with point (4, 1): $y = -\frac{1}{3}x + \frac{7}{3}$

Second formula is:

Lower curve: y = 1
Upper curve: $y = -\frac{1}{3}x + \frac{7}{3}$
With bounds: $1 \leq x \leq 4$

Integrals should be split into two parts.

Type II Region

The right and left curves are required, with $1 \leq y \leq 2$. Need to express x as a function of y.
Curve equations:

1. Left line: $y = x + 1 \implies x = y - 1$

2. Right line: $y = -\frac{1}{3}x + \frac{7}{3} \implies x = 7 - 3y$

Limits of integration:

• y: 1 to 2
• x: y-1 to 7-3y

Integral Setup

$\int<em>{1}^{2} \int</em>{y-1}^{7-3y} y^2 dx dy$

• Initial integration with respect to x results to:

$\int<em>{1}^{2} y^2[x]</em>{y-1}^{7-3y} dy = \int_{1}^{2} y^2[(7-3y) - (y-1)] dy$

Simplifying leads to:
$\int<em>{1}^{2} y^2 (8 - 4y) dy = \int</em>{1}^{2} (8y^2 - 4y^3) dy$

Compute antiderivative and apply limits:

Antiderivative: $\frac{8y^3}{3} - y^4$
With bounds 1 to 2:

$\frac{8(2)^3}{3} - (2)^4 - (\frac{8(1)^3}{3} - (1)^4) = \frac{64}{3} - 16 - \frac{8}{3} + 1$

Simplify to:

$(\frac{64}{3} - \frac{8}{3}) - (16 - 1) = \frac{56}{3} - 15 = \frac{56 - 45}{3} = \frac{11}{3}$

The final result is $\frac{11}{3}$.

Triple Integration: Setting Up the Integral

Region E

• Partition each axis (x, y, z).
• Create mini boxes within the larger region.

Micro Box

• A small box with sides $\,\,\,\,\,\,\,\,\,{\Delta}x$, $\,\,\,\,\,\,\,\,\,{\Delta}y$, and $\,\,\,\,\,\,\,\,\,{\Delta}z$.
• Denoted as $B_{ijk}$, defined by:
  • $[x<em>{i-1}, x</em>i] \times [y<em>{j-1}, y</em>j] \times [z<em>{k-1}, z</em>k]$

Sample Point

• Choose a sample point $(x<em>{ijk}^<em>, y</em>{ijk}^</em>, z_{ijk}^*)$ within each micro-box.
• Evaluate the function f at this point: $f(x<em>{ijk}^<em>, y</em>{ijk}^</em>, z_{ijk}^*)$.

Riemann Sum

• Approximate the triple integral using a triple Riemann sum:

$\sum<em>{i=1}^{l} \sum</em>{j=1}^{m} \sum<em>{k=1}^{n} f(x</em>{ijk}^<em>, y<em>{ijk}^</em>, z</em>{ijk}^*) {\Delta}V$

where ${\Delta}V = {\Delta}x {\Delta}y {\Delta}z$.

Existence and Computation

• If f is continuous on the box, the limit of the Riemann sum exists as l, m, and n approach infinity.
• The computation method is what comes down to Fubini's theorem.

Fubini's Theorem

• Allows iterated integrals to compute the triple integral:

$\int \int \int<em>{Box} f(x, y, z) dV = \int</em>{r}^{s} \int<em>{c}^{d} \int</em>{a}^{b} f(x, y, z) dx dy dz$

• The order of integration (dx, dy, dz) can be permuted, resulting in six possible iterated integrals.

Calculation

Compute $\int \int \int_{E} (x + yz^2) dV$

with E defined as

• z from 0 to 1
• y from 2 to 4
• x from -1 to 5

Setup

$\int<em>{0}^{1} \int</em>{2}^{4} \int_{-1}^{5} (x + yz^2) dx dy dz$

Process:

• First integration: integrate with respect to x:
  *Remainder: integrate with respect to y, and apply the limits from 2 to 4, then integrate the y variable to the remainder and apply limits.
• Apply last limits, after having only z as a variable to the rest functions. The result is 36