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Question #41187

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be the area of the region in the $xy$ -plane bounded by the ellipse $4x^2 + 9y^2 = 36$ , the positive $x$ -axis, and the ray originating from the origin with polar angle $\phi = \frac{5\pi}{6}$ . Find the exact value of $A$ .

Options

A.
$3\\pi - 3\\arctan\\left(\\frac{\\sqrt{3}}{2}\\right)$
B.
$3\\arctan\\left(\\frac{\\sqrt{3}}{2}\\right)$
C.
$\\frac{3\\pi}{2} - 3\\arctan\\left(\\frac{\\sqrt{3}}{2}\\right)$
D.
$3\\pi + 3\\arctan\\left(\\frac{\\sqrt{3}}{2}\\right)$

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

Algebraic-Trigonometric Graphical Mapping|Ellipse Auxiliary Circle|The Quadrant Adjustment Logic

Why This Question Is Tricky

Common Mistakes

This question tests the 'Quadrant Adjustment Logic' when mapping between the polar angle and the eccentric angle of an ellipse.

A common trap is evaluating the sector area using the polar integral $A = \\frac{1}{2} \\int r^2 d\\phi$ , which yields an antiderivative involving $\\arctan(\\frac{a}{b}\\tan\\phi)$ .

Students often blindly apply the Fundamental Theorem of Calculus across the boundary $\\phi = \\pi/2$ , ignoring the discontinuity of the principal arctan branch.

This results in an incorrect negative area or a missing quadrant area (Option B or C).

The correct approach requires either splitting the integral at the discontinuity or geometrically mapping to the eccentric angle while strictly enforcing the quadrant constraint $\\theta \\in (\\pi/2, \\pi)$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Algebraic-Trigonometric Graphical Mapping
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free