MAT291: Lecture 13 - Triple Integrals

Overview

Parinaz Naseri extends double-integral ideas to triple integrals, explaining volume-based Riemann sums, iterated integration, and why Fubini’s theorem permits six orders without guaranteeing equal difficulty. Worked examples evaluate a rectangular-box integral as 48 and a trigonometric integral as 4/π, then develop bounds for a first-octant tetrahedron and integrate xy over it to obtain 32/15.

Key takeaways

A triple integral is the limit of sums of function values multiplied by small box volumes ΔxΔyΔz, extending the Riemann-sum interpretation of double integrals.
For a rectangular box, integrating 1 gives its volume, while the average value of f is ∭f dV divided by the region’s volume.
Fubini’s theorem allows six integration orders for three variables, but choosing an order that avoids difficult antiderivatives can greatly simplify the calculation.
For the integrand xz cos(xy) over 0≤x≤1, 0≤y≤π, and 0≤z≤2, integrating y or z before x gives 4/π; integrating x first introduces integration by parts.
For the tetrahedron 2x+y+z≤4 in the first octant, the order dz dy dx gives bounds 0≤z≤4−2x−y, 0≤y≤4−2x, and 0≤x≤2.
The integral of xy over that tetrahedron is 32/15; accurately sketching the solid and deriving its variable-dependent bounds is the central setup challenge.

Chapters

0:00 Triple Integrals as Riemann Sums of Small 3D Boxes

A triple integral extends the double-integral idea from area elements to small boxes with dimensions Δx, Δy, and Δz.
For each box, evaluate f(x,y,z) at a point and multiply by the box’s volume; summing over the region and refining the boxes yields the integral.
A triple integral describes a four-dimensional graph in general, so the computation uses spatial volume elements rather than a visualization of the full graph.

5:30 Volume and Average Value over a 3D Region

Integrating 1 over a box gives its volume; for bounds a≤x≤b, c≤y≤d, and e≤z≤f, the volume is (b−a)(d−c)(f−e).
The average value of f over a region is its triple integral divided by the region’s volume.
Iterated integration proceeds from the innermost integral outward, with each step leaving a function of the variables not yet integrated.

7:13 Rectangular-Box Example: Integrating 6x²y + 4z

Naseri evaluates ∭(6x²y+4z) dV over 0≤x≤1, 0≤y≤2, and 0≤z≤3, using dz dy dx.
In the z integral, x and y are constants, so integrating gives 18x²y+18.
Integrating over y and then x gives the final value 48; reversing the integration order should produce the same result.

12:41 Choosing an Efficient Order for xz cos(xy)

For xz cos(xy) on 0≤x≤1, 0≤y≤π, and 0≤z≤2, integrating with respect to y or z first avoids unnecessary integration by parts.
Keeping x for the innermost integral is the difficult choice because integrating x cos(xy) with respect to x requires integration by parts.
Fubini’s theorem allows all six orders of dx, dy, and dz, but the integrand’s form can make some orders substantially easier.

15:15 Two Easy Orders Both Give 4/π

With dz first, integrating z from 0 to 2 reduces the integrand to 2x cos(xy); integrating y then produces 2 sin(πx).
The remaining x integral evaluates to 4/π.
With dy first, the y integral gives z sin(πx); integrating z from 0 to 2 leads to the same result, 4/π.

21:26 Why Integrating x First Requires Integration by Parts

Taking x as the innermost variable requires integrating x cos(xy) while treating y as a constant.
Using integration by parts with u=x and dv=cos(xy) dx gives v=sin(xy)/y and leaves another integral to evaluate.
This example shows why choosing a convenient order can reduce work even though all six orders are mathematically valid.

25:20 Simple Solid Regions and Bounds That Depend on Other Variables

A simple 3D region can be described by bounding one variable between two functions of the other two, then bounding another variable in terms of the remaining one.
For an order dz dy dx, z is typically bounded by surfaces depending on x and y, y by curves depending on x, and x by constants.
Changing the integration order requires finding a new description of the same solid; the bounds do not simply stay unchanged.

30:29 Sketching the Tetrahedron and Deriving Its Bounds

The solid is bounded by x=0, y=0, z=0, and the plane 2x+y+z=4, whose axis intercepts are x=2, y=4, and z=4.
Solving the plane equation for z gives 0≤z≤4−2x−y.
Projecting onto the xy-plane gives 0≤y≤4−2x and 0≤x≤2; z=4 is only the maximum height, not the upper bound for every point.

39:20 Evaluating the Tetrahedron Integral of xy

For ∭S xy dV, the bounds are integrated in the order dz dy dx: z from 0 to 4−2x−y, y from 0 to 4−2x, and x from 0 to 2.
After the z integration, setting A=4−2x simplifies the y integral to x∫₀ᴬ(Ay−y²)dy=xA³/6.
Substituting u=4−2x in the final integral yields ∫S xy dV=32/15.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Parinaz Naseri.