## Calculus 8th Edition

$=\dfrac{27}{8}$
In polar co-ordinates, we have $r^2=x^2+y^2 \implies r=\sqrt{x^2+y^2}$ and $x=r \cos \theta \\ y=r \sin \theta$ Thus, $I= \int_{0}^{ \pi/2}\int_0^3 \int_0^{1/3} [z dx] r dr d\theta=\int_{0}^{ \pi/2}\int_0^3 (z) [x]_0^{1/3} r dr d\theta$ and $\int_{0}^{ \pi/2}\int_0^3 r \sin \theta[(\dfrac{1}{3}) r \cos \theta]r dr d\theta=\int_{0}^{ \pi/2}\int_0^3 (\dfrac{r^3}{3}) \sin \theta \cos \theta dr d\theta$ or, $(\dfrac{1}{12}) \int_{0}^{ \pi/2}[r^4]_0^3 \sin \theta \cos \theta d\theta=\dfrac{81}{12} \int_{0}^{ \pi/2} \sin \theta \cos \theta d\theta$ Now consider$p=\sin \theta$ and $d\theta=\cos \theta dp$ $\dfrac{81}{12} \int_{0}^{ \pi/2} p dp=\dfrac{81}{12} [p^2/2]_{0}^{ \pi/2} =\dfrac{27}{8}$