1.23. Compute the triple integral:

a) $\iiint_{\Omega}(2 x+3 y+z) d x d y d z$, where $\Omega=\{(x, y, z): x \geqslant 0 ; y \geqslant 0$; $0 \leqslant z \leqslant 3 ; x+y \leqslant 3\}$

b) $\iiint_{\Omega}(1-x)^{3} \sqrt{1-y^{2}} d x d y d z$, where $\Omega=\{(x, y, z):|x| \leqslant 1 ;|y| \leqslant 1$; $|z| \leqslant 3\}$

c) $\iiint_{\Omega} \frac{x y}{\sqrt{z}} d x d y d z$, where $\Omega=\left\{(x, y, z): z^{2} \leqslant x^{2}+y^{2} ; x \geqslant 0 ; y \geqslant 0\right.$; $0 \leqslant z \leqslant 1\}$
Solution.

a) We reduce the calculation of this integral to a triple repeated integration:

$$
\begin{gathered}
\iiint_{\Omega}(2 x+3 y+z) d x d y d z=\int_{0}^{4} d x \int_{0}^{4-x} d y \int_{0}^{3}(2 x+3 y+z) d z= \\
=\left.\int_{0}^{4} d x \int_{0}^{4-x}\left(2 x z+3 y z+\frac{z^{2}}{2}\right)\right|_{0} ^{3} d y=\int_{0}^{4} d x \int_{0}^{4-x}\left(6 x+9 y+\frac{9}{2}\right) d y= \\
=\left.\int_{0}^{4}\left(6 x y+\frac{9 y^{2}}{2}+\frac{9}{2} y\right)\right|_{0} ^{4-x} d x= \\
=\int_{0}^{4}\left(6 x(4-x)+\frac{9}{2}\left((4-x)^{2}+4-x\right)\right) d x=\int_{0}^{4}\left(-\frac{3}{2} x^{2}-\frac{33}{2} x+144\right) d x= \\
=\left.\left(-\frac{x^{3}}{2}-\frac{33 x^{2}}{4}+144 x\right)\right|_{0} ^{4}=-32-132+576=412
\end{gathered}
$$

b) In this case, the region $\Omega$ is a parallelepiped. We set the limits and perform repeated integration. Essentially, it all boils down to the product of three independent single integrals:

$$
\begin{gathered}
\iiint_{\Omega}(1-x)^{3} \sqrt{1-y^{2}} d x d y d z=\int_{-1}^{1}(1-x)^{3} d x \int_{-1}^{1} \sqrt{1-y^{2}} d y \int_{-3}^{3} d z= \\
=\left.\left(-\frac{(1-x)^{4}}{4}\right)\right|_{-1} ^{1} \cdot \int_{-1}^{1} \sqrt{1-y^{2}} d y \cdot\left(\left.z\right|_{0} ^{3}\right)=12 \pi
\end{gathered}
$$

The middle integral is numerically equal to $\frac{\pi}{2}$, the area of half a circle $x^{2}+y^{2} \leqslant 1, y \geqslant 0$.

c) For solving the problem, it is convenient to switch to cylindrical coordinates $x=\rho \cos \varphi ; y=\rho \sin \varphi ; z=z$ with the Jacobian

$$
J=\frac{D(x, y, z)}{D(\varphi, \rho, z)}=\rho
$$

In this case, the region $\Omega$ transforms into the region

$$
\begin{gathered}
D=\left\{(\varphi, \rho, z): 0 \leqslant \varphi \leqslant \frac{\pi}{2} ; \quad 0 \leqslant z \leqslant \rho \leqslant 1\right\} \\
\iiint_{\Omega} \frac{x y}{\sqrt{z}} d x d y d z=\iiint_{D} \frac{\rho^{3} \cos \varphi \sin \varphi}{\sqrt{z}} d \varphi d \rho d z=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin 2 \varphi d \varphi \int_{0}^{1} \rho^{3} d \rho \int_{0}^{\rho} \frac{d z}{\sqrt{z}}= \\
=\left.\left.\left(-\frac{1}{4} \cos 2 \varphi\right)\right|_{0} ^{\frac{\pi}{2}} \cdot \int_{0}^{1} 2 \rho^{3}(\sqrt{z})\right|_{0} ^{\rho} d \rho=\int_{0}^{1} \rho^{\frac{7}{2}} d \rho=\left.\frac{2}{9} \rho^{\frac{9}{2}}\right|_{0} ^{1}=\frac{2}{9} .
\end{gathered}
$$