Problem:
Determine the volume of the region that lies behind the plane $xyz=8$ and in front of the region in the $yz$-plane that is bounded by $z=\frac{3}{2}\sqrt{y}$ and $z=\frac{3}{4}y$.

Solution:

To find the volume of the region, we need to calculate the difference between the upper and lower boundaries of the region in the $xyz$-space.

Finding the upper boundary:
The upper boundary of the region is given by the plane $xyz=8$. To find the equation of this plane, we need to solve for $z$ in terms of $x$ and $y$:

$xyz=8$

$z=\frac{8}{xy}$

Therefore, the equation of the upper boundary is $z=\frac{8}{xy}$.

Finding the lower boundary:
The lower boundary of the region is given by the region in the $yz$-plane that is bounded by $z=\frac{3}{2}\sqrt{y}$ and $z=\frac{3}{4}y$. To find the equation of this region, we need to find the intersection points of these two surfaces.

Setting the two equations equal to each other, we have:

$\frac{3}{2}\sqrt{y}=\frac{3}{4}y$

Squaring both sides, we get:

$\frac{9}{4}y=\frac{9}{16}y^2$

Multiplying both sides by $\frac{16}{9}$, we have:

$4y=y^2$

Rearranging the equation, we get:

$y^2-4y=0$

Factoring out $y$, we have:

$y(y-4)=0$

Therefore, $y=0$ or $y=4$.

Substituting these values into the equation $z=\frac{3}{2}\sqrt{y}$, we get:

$z=\frac{3}{2}\sqrt{0}=0$

$z=\frac{3}{2}\sqrt{4}=3$

Therefore, the equation of the lower boundary is $0 \leq z \leq 3$.

Calculating the volume:
To calculate the volume, we integrate the difference between the upper and lower boundaries with respect to $x$, $y$, and $z$.

$\int\int\int (z_{\text{upper}}-z_{\text{lower}}) \, dx \, dy \, dz$

$\int\int\int (\frac{8}{xy}-z) \, dx \, dy \, dz$

Since the region is bounded by the planes $x=0$, $y=0$, and $z=0$, we integrate with respect to $x$, $y$, and $z$ over the appropriate ranges.

$\int_0^8 \int_0^4 \int_0^3 (\frac{8}{xy}-z) \, dz \, dy \, dx$

Simplifying the integral, we have:

$\int_0^8 \int_0^4 (\frac{8}{xy}z-\frac{1}{2}z^2) \, dy \, dx$