Problem:

Determine the volume of the region that lies behind the plane x y z=8 and in front of the region in the yz- plane that is bounded by z=3/2√y and z=3/4 y is?

Solution:

Step 1: Sketch the region in the yz-plane bounded by z=3/2√y and z=3/4 y.

Step 2: Determine the range of y values for which the region in the yz-plane exists.

Step 3: Express the equation of the plane x y z=8 in terms of y and z.

Step 4: Determine the range of z values for which the region behind the plane x y z=8 exists.

Step 5: Determine the volume of the region by integrating the function 1 with respect to y and z over the ranges determined in steps 2 and 4.

The detailed solution is as follows:

Step 1: Sketch the region in the yz-plane bounded by z=3/2√y and z=3/4 y.

The region in the yz-plane bounded by z=3/2√y and z=3/4 y is shown below:


Step 2: Determine the range of y values for which the region in the yz-plane exists.

The region in the yz-plane exists for y between 0 and 16.

Step 3: Express the equation of the plane x y z=8 in terms of y and z.

The equation of the plane x y z=8 can be written as z=8/x y. Therefore, in terms of y and z, the equation of the plane is z=8/y.

Step 4: Determine the range of z values for which the region behind the plane x y z=8 exists.

The region behind the plane x y z=8 exists for z between 0 and 8/y.

Step 5: Determine the volume of the region by integrating the function 1 with respect to y and z over the ranges determined in steps 2 and 4.

The volume of the region is given by the double integral:

V = ∫∫R 1 dy dz

where R is the region of integration.

The limits of integration are:

0 ≤ y ≤ 16

0 ≤ z ≤ 3/4 y

8/y ≤ z ≤ 8

Therefore, the volume is given by:

V = ∫0^16 ∫8/y^(3/2)^(3/4 y) dy dz + ∫0^16 ∫3/4 y^8/y dz dy

Simplifying the integral, we get:

V = 64/15

Therefore, the volume of the region is 64/15.