Question

For a continuous function \(f(r, \theta)\) defined on a region D in \((r, \theta)\) plane, the area of the domain D is given by:

• A. \(\iint_D f(r, \theta) \, dr \, d\theta\)
• B. \(\iint_D r f(r, \theta) \, dr \, d\theta\)
• C. \(\iint_D r \, dr \, d\theta\)
• D. \(\iint_D dr \, d\theta\)

Hint:

Exam Tip:
In polar coordinates:

• \(x = r \cos \theta\), \(y = r \sin \theta\).
• The Jacobian is \(r\).
• \(dA = r \, dr \, d\theta\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to find the formula for the area of a region expressed in polar coordinates.

Step 2: Key Formula or Approach:
In polar coordinates, the area element is \(dA = r \, dr \, d\theta\).
So, the area of a region D is: \[ \text{Area} = \iint_D dA = \iint_D r \, dr \, d\theta \]

Step 3: Detailed Explanation:
The options are:
(A) \(\iint_D f(r, \theta) \, dr \, d\theta\) — This would be an integral of \(f\) without the Jacobian.
(B) \(\iint_D r f(r, \theta) \, dr \, d\theta\) — This would be the integral of \(r f\), not area.
(C) \(\iint_D r \, dr \, d\theta\) — This is the correct area element.
(D) \(\iint_D dr \, d\theta\) — This is missing the Jacobian \(r\).
So, the area is given by option (C).

Step 4: Final Answer:
Therefore, option (C) is correct.