Questions: Triple Integrals in Cartesian Coordinates

Question 1

You are integrating over the solid tetrahedron with vertices at (0,0,0), (1,0,0), (0,1,0), and (0,0,1), using the order dz dy dx. What is the correct upper bound for z in the innermost integral?

• A. z = 1, since z ranges from 0 to 1 overall in the region
• B. z = 1 − x, since the region's z-extent depends only on x
• C. z = 1 − x − y, since the slanted face satisfies the plane x + y + z = 1
• D. z = √(1 − x² − y²), the boundary of a hemisphere

Answer: (C)

The tetrahedron is bounded above by the plane x + y + z = 1, so z ≤ 1 − x − y for any fixed (x, y) inside the region. Using z = 1 as the upper bound (option A) is the most common error — it ignores that the ceiling on z depends on where you are in the xy-plane. Option B is also wrong; the bound depends on both x and y. This is the essential challenge of triple integrals: inner bounds are generally functions of the outer variables.

Question 2

Which statement about changing the order of integration in a triple integral is correct?

• A. The integral's value changes when you reorder because different bounds are used
• B. You can freely swap any two variables without rederiving the bounds
• C. The value is unchanged by Fubini's theorem, but the bounds must be completely rederived for the new ordering
• D. Reordering is only valid if the integration region is a rectangular box

Answer: (C)

Fubini's theorem guarantees that any ordering of integration yields the same value — but only if you correctly re-describe the region W for the new ordering. The bounds for one ordering (e.g., the z bounds as a function of x and y) are completely different from the bounds for another ordering (the x bounds as a function of y and z). Reordering is not just swapping variable names — it requires redrawing and re-reading the geometry of the region from a new perspective.

Question 3

For the unit cube [0,1]³, all six possible orderings of integration produce the same result with the same constant bounds [0,1] for each variable.

• T. True
• F. False

Answer: True

The unit cube has constant bounds in every direction: x ∈ [0,1], y ∈ [0,1], z ∈ [0,1], independent of the other variables. So no matter what order you integrate, the bounds are always the same constants and the value is the same. This is the special simplicity of rectangular regions — the bounds decouple. For any non-rectangular region, the bounds of inner variables will depend on outer variables, so reordering requires re-deriving them.

Question 4

For a non-rectangular region in 3D (like a tetrahedron or a ball), you can set the bounds for each variable independently of the other variables.

• T. True
• F. False

Answer: False

For non-rectangular regions, inner bounds must account for the shape of the region at each fixed value of the outer variables. For example, integrating over the unit ball x² + y² + z² ≤ 1 in order dz dy dx, the z bounds are −√(1−x²−y²) to √(1−x²−y²) — depending on both x and y. Using fixed bounds [−1, 1] for all three variables would integrate over the cube containing the ball, not the ball itself, giving the wrong answer.

Question 5

Why might you choose to change the order of integration in a triple integral even though the final value is the same regardless of order?

Model answer: Different orderings produce different inner integrands, some of which may be far harder to evaluate analytically than others. For example, one ordering might require integrating e^(z³) with respect to z (which has no elementary antiderivative), while a different ordering avoids this entirely. By Fubini's theorem the value is the same, but choosing the right order can mean the difference between a tractable computation and one that cannot be done in closed form.

This is the practical payoff of understanding Fubini's theorem deeply. Recognizing that order can be changed — and that the key work is re-describing the region's bounds for the new order — is what allows you to turn impossible integrals into manageable ones. Sketching the region and asking 'which variable is easiest to integrate last?' (as the outermost integral) is a standard strategic move in multivariable calculus.