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Question #40880

Difficulty: 8/10 | 🟢 Free

The Question

Let VV be the region in R3\mathbb{R}^3 defined by ∣x∣+∣y∣+∣z∣≤2|x| + |y| + |z| \le 2. Evaluate the integral: I=∭V(x2{x}+y2{y}+z2{z}) dx dy dzI = \iiint_V \left( x^2 \{x\} + y^2 \{y\} + z^2 \{z\} \right) \,dx\,dy\,dz where {t}\{t\} denotes the fractional part of tt.

Options

  • A.

    325\frac{32}{5}

  • B.

    645\frac{64}{5}

  • C.

    0

  • D.

    −325-\frac{32}{5}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Octant Sign Convention|Axis of Integration Selection|Integral Part and Fractional Part

Why This Question Is Tricky

Common Mistakes

This question tests the evaluation of volume integrals involving the fractional part function over a symmetric 3D region.

The primary trap lies in the parity of the fractional part function {x}\{x\}.

Students frequently and incorrectly assume that {x}\{x\} is an odd function, which would lead them to erroneously conclude that the integral evaluates to zero over the symmetric domain.

Furthermore, when converting {x}\{x\} to x−[x]x - [x], students often mishandle the greatest integer function for negative values, forgetting that [−x]=−1−[x][-x] = -1 - [x] for non-integers.

The problem elegantly rewards the application of the identity ∫−aaf(x)[x]dx=−∫0af(x)dx\int_{-a}^a f(x)[x]dx = -\int_0^a f(x)dx for even functions, bypassing tedious piecewise integration across different octants while rigorously testing the sign convention of step functions.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntroduction to Three-dimensional Geometry - Class 11
ConceptOctant Sign Convention
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree