Discussion
The area of the circle \( r = a \) by double integration is
Answer: C
Integrate the polar area element over the whole disc:
\[ A \;=\; \int_{0}^{2\pi}\!\!\int_{0}^{a} r\,dr\,d\theta \;=\; \int_{0}^{2\pi}\frac{a^{2}}{2}\,d\theta \;=\; \frac{a^{2}}{2}(2\pi) \;=\; \pi a^{2}. \]
Dropping the factor \( r \) would give \( 2\pi a \), which is the circumference and not an area at all
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