Discussion
Integrating \( r \) over the disc \( r \le a \) in polar coordinates gives
Answer: B
The integrand is \( r \) and the element supplies another factor of \( r \):
\[ \int_{0}^{2\pi}\!\!\int_{0}^{a} r \cdot r\,dr\,d\theta \;=\; 2\pi\left[\frac{r^{3}}{3}\right]_{0}^{a} \;=\; \frac{2\pi a^{3}}{3}. \]
Two powers of \( r \) appear, one from the integrand and one from the area element, which is what raises the result to a cube
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