How can we calculate the volume of the solid that lies under the plane defined by the equation 3x + 2y + z = 12 and above the rectangle defined by the bounds 0 ≤ x ≤ 1 and 3 ≤ y ≤ 4?

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To find the volume of the solid that lies beneath the plane 3x + 2y + z = 12 and above the rectangle R defined by 0 ≤ x ≤ 1 and 3 ≤ y ≤ 4, follow these steps:

Step 1: Rearrange the Plane Equation

First, we rearrange the equation of the plane to express z in terms of x and y:

z = 12 - 3x - 2y

Step 2: Define the Volume Integral

The volume V under the plane and above the rectangle can be calculated using a double integral:

V = ∬_R z \, dA

where dA is the area element in the xy-plane.

Step 3: Set Up the Double Integral

Here, the bounds for x range from 0 to 1, and for y, they range from 3 to 4. Thus, we set up the integral:

V = ∫[0, 1] ∫[3, 4] (12 - 3x - 2y) \, dy \, dx

Step 4: Evaluate the Inner Integral

We first evaluate the inner integral with respect to y:

∫[3, 4] (12 - 3x - 2y) \, dy = [12y - 3xy - y^2] |_[3, 4]

Calculating this gives:

=[(12(4) - 3x(4) - (4)^2) - (12(3) - 3x(3) - (3)^2)]
= [(48 - 12x - 16) - (36 - 9x - 9)]
= (32 - 12x) - (27 - 9x) = 5 - 3x

Step 5: Evaluate the Outer Integral

Now, substitute back into the outer integral:

V = ∫[0, 1] (5 - 3x) \, dx

This evaluates to:

=[5x - (3/2)x^2] |_[0, 1] = [5(1) - (3/2)(1)^2] - [0] = 5 - 1.5 = 3.5

Conclusion

The volume of the solid that lies under the plane 3x + 2y + z = 12 and above the rectangle R is:

V = 3.5 cubic units

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