---
title: "Triple integral of log z"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:triple-integral-log-z"
source: https://www.bctengineer.com/q/semester_2/mathematics/triple-integral-log-z
site: BCT Engineering
---

# Triple integral of log z

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2069 Poush, 5 marks
- Video: [TRIPLE INTEGRAL Evaluate ∫∫∫ log z dydxdz](https://www.youtube.com/watch?v=n3yXXAhx3pA) — m-easy maths
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/triple-integral-log-z
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Evaluate ∫₁ᵉ ∫₁^(log y) ∫₁^(eˣ) log z dz dx dy.

## Short answer

After integrating log z, then x, then y, the result is I = (e² - 8e + 13)/4 ≈ -0.34.

## Answer

I = ∫₁ᵉ ∫₁^{log y} ∫₁^{eˣ} log z dz dx dy

The integration order is z (innermost), then x, then y.

#### z-integration
∫ log z dz = z log z − z, so

```text
∫₁^{eˣ} log z dz = [z log z − z]₁^{eˣ} = (x eˣ − eˣ) − (0 − 1) = (x − 1)eˣ + 1
```

#### x-integration (1 to log y)
∫ (x − 1)eˣ dx = (x − 2)eˣ, so

```text
∫₁^{log y} [(x−1)eˣ + 1] dx = [(x − 2)eˣ + x]₁^{log y}
  = (log y − 2) y + log y − (−e + 1)
  = y log y − 2y + log y + e − 1
```

#### y-integration (1 to e)
```text
∫₁ᵉ y log y dy = [ (y²/2) log y − y²/4 ]₁ᵉ = e²/2 − e²/4 + 1/4 = e²/4 + 1/4
∫₁ᵉ 2y dy      = e² − 1
∫₁ᵉ log y dy   = [y log y − y]₁ᵉ = 0 + 1 = 1
∫₁ᵉ (e − 1) dy = (e − 1)²
```

```text
I = (e²/4 + 1/4) − (e² − 1) + 1 + (e − 1)²
  = e²/4 + 13/4 − 2e
  = (e² − 8e + 13)/4
```

**Answer: I = (e² − 8e + 13)/4 ≈ −0.34.**

The value is negative because for 1 < y < e the upper limit log y of x is below the lower limit 1, so that part is integrated backwards. The integral is evaluated exactly as printed.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the result of the z-integration of log z from 1 to e^x?
   A. (x - 1)e^x
   B. x e^x - e^x - 1
   C. (x - 1)e^x + 1
   D. x e^x + 1
   Answer: C. Using the antiderivative z log z - z, the upper limit gives x e^x - e^x and the lower limit gives -1, so the value is (x - 1)e^x + 1. (Key point: Innermost z-integral)

2. What is the integral of (x - 1)e^x with respect to x?
   A. (x - 2)e^x
   B. (x - 1)e^x
   C. x e^x
   D. (x^2/2 - x)e^x
   Answer: A. By parts, the integral of x e^x is (x - 1)e^x and the integral of e^x is e^x, so the result is (x - 2)e^x. (Key point: Integration by parts in x)

3. Why does the answer come out negative (about -0.34)?
   A. The function log z is negative on the range of z
   B. The y-integral has its limits reversed
   C. The z-integral always gives a negative value
   D. For 1 < y < e the upper limit log y of x is below the lower limit 1, so the x-integral runs backwards
   Answer: D. log y lies between 0 and 1 for 1 < y < e, below the x lower limit 1, so that part is integrated backwards. (Key point: Why the value is negative)

4. What is the final value of the integral?
   A. (e^2 + 8e + 13)/4
   B. (e^2 - 8e + 13)/4
   C. (e^2 - 8e - 13)/4
   D. (e^2 - 4e + 13)/4
   Answer: B. Adding (e^2/4 + 1/4) - (e^2 - 1) + 1 + (e - 1)^2 gives e^2/4 + 13/4 - 2e = (e^2 - 8e + 13)/4. (Key point: Final value)

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