Discussion
The double integral of unity over the disc \( x^{2} + y^{2} \le 9 \) equals
Answer: C
An integrand of one returns the area, and in polar form
\[ \int_{0}^{2\pi}\!\!\int_{0}^{3} r\,dr\,d\theta \;=\; 2\pi\left[\frac{r^{2}}{2}\right]_{0}^{3} \;=\; 2\pi\left(\frac{9}{2}\right) \;=\; 9\pi. \]
This is \( \pi r^{2} \) with \( r = 3 \). Taking the radius as nine rather than its square would give \( 81\pi \)
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