9.1. calcula $iiint_s z sqrt{x^2 + y^2} dv$, donde $s$ es el sólido de base semicircular de radio 1 y altura…

Explanation:

Step1: Convert to cylindrical coordinates

In cylindrical coordinates, (x = r\cos\theta), (y = r\sin\theta), (z=z) and (dV=r\ dz\ dr\ d\theta), and (\sqrt{x^{2}+y^{2}}=r). The base is a semicircle with (r\in[0,1]), (\theta\in[0,\pi]) and (z\in[0,2]). The integral becomes (\int_{0}^{\pi}\int_{0}^{1}\int_{0}^{2}z\cdot r\cdot r\ dz\ dr\ d\theta=\int_{0}^{\pi}\int_{0}^{1}\int_{0}^{2}r^{2}z\ dz\ dr\ d\theta)

Step2: Integrate with respect to (z)

(\int_{0}^{\pi}\int_{0}^{1}r^{2}\left[\frac{z^{2}}{2}\right]{0}^{2}dr\ d\theta=\int{0}^{\pi}\int_{0}^{1}r^{2}\cdot\frac{4 - 0}{2}dr\ d\theta=\int_{0}^{\pi}\int_{0}^{1}2r^{2}dr\ d\theta)

Step3: Integrate with respect to (r)

(\int_{0}^{\pi}2\left[\frac{r^{3}}{3}\right]{0}^{1}d\theta=\int{0}^{\pi}2\cdot\frac{1 - 0}{3}d\theta=\frac{2}{3}\int_{0}^{\pi}d\theta)

Step4: Integrate with respect to (\theta)

(\frac{2}{3}[\theta]_{0}^{\pi}=\frac{2}{3}(\pi - 0))

Answer:

(\frac{2\pi}{3})