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Question #41662

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the surface of revolution generated by revolving the curve $y = \sin(2x) \sqrt{|\cos(2x)|}$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ about the $x$ -axis. A variable plane perpendicular to the $x$ -axis with $x$ -intercept $c$ cuts $S$ to form a circular cross-section of area $A(c)$ . Evaluate the integral: $I = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \frac{A(c) \sec^2 c}{1 + e^{\tan^3 c}} \, dc$

Options

A.
2
B.
$-\frac{\pi}{2}$
C.
$1 - \frac{\pi}{4}$
D.
$\frac{\pi}{2}$

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

Variable plane intercepts|Substitution (Change of Variable)|The Sign-Consistency Principle|Even and Odd Function Symmetry

Why This Question Is Tricky

Common Mistakes

This question tests the Sign-Consistency principle and symmetry properties in definite integrals.

Students must first correctly formulate the cross-sectional area, recognizing that squaring the function preserves the absolute value $|\cos(2x)|$ .

A common trap is mechanically applying the even/odd property $\int_{-a}^a \frac{f(x)}{1+e^{g(x)}} dx = \int_0^a f(x) dx$ but failing to split the resulting integral at $c = \pi/4$ where the sign of $\cos(2c)$ flips.

Ignoring the absolute value leads to the erroneous evaluation of $-\pi/2$ , while failing to recognize the even/odd symmetry might cause students to abandon the problem due to the intimidating $e^{\tan^3 c}$ term.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Variable plane intercepts
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free