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Question #68992

Difficulty: 8/10 | 🟢 Free

The Question

Let $R$ be the region in the complex plane defined by $z = x+iy$ such that $0 \le x \le 2$ and $\min(x, x^2) \le y \le \max(x, x^2)$ . Let $R'$ be the image of the region $R$ under the transformation $w = \bar{z}^2$ . Let $A$ be the area of the region $R'$ , and let $I = \iint_{R'} \frac{\text{Re}(w)}{|w|} du dv$ , where $w = u+iv$ . Find the value of $3(A + I)$ .

Options

A.
60
B.
116
C.
32
D.
144

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Axis of Integration Selection|Compound Function Boundaries|Locus Problems via Transformation|The Sign-Consistency Principle

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

The core trap is the 'Sign-Consistency Principle' in multivariable transformations.

When mapping a region via an orientation-reversing transformation like $w = \bar{z}^2$ , the Jacobian determinant is negative ( $J = -4|z|^2$ ).

Students frequently substitute $du dv = J dx dy$ directly into the integral $I$ without taking the absolute value $|J|$ , assuming the algebraic substitution preserves the area's positivity.

This leads to an inverted sign for $I$ (calculating it as $+28/3$ instead of $-28/3$ ), yielding the trap answer 116.

Furthermore, the compound boundary $\min(x,x^2)$ to $\max(x,x^2)$ forces students to correctly split the integration axis, punishing those who blindly integrate from $x^2$ to $x$ over the entire interval $[0,2]$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Axis of Integration Selection
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free