Find the volume of the solid that lies under the plane 3x + 2y + z = 12, 3x + 2y + z = 12 and above the rectangle R = {(x, y) |0 ≤ x ≤1, - 3 ≤ y ≤ 4}

Solution:

Step 1:

Consider z = 12 - 3x - 2y

With the limits 0 ≤ x ≤ 1, - 3 ≤ y ≤ 4

Step 2:-

We do double integration to find the volume under the plane.

The volume of the solid under the given plane is \(\int \int z \,dA =\int \int f(x,y). dA\).

\(v=\int_{0}^{1}\int_{-3}^{4}(z dy.dx)\\\\v=\int_{0}^{1}\int_{-3}^{4}(12 - 3x - 2y) dy.dx \\\\=\int_{0}^{1}\int_{-3}^{4}((12 - 3x - 2y) dy.dx )\\\\=\int_{0}^{1}(12 [y]_{-3}^{4} - 3x [y]_{-3}^{4} - [y^2]_{-3}^{4}) dx\\\\= \int_{0}^{1}[84 - 21x -7]dx\\\\=\int_{0}^{1}[77 -21x] dx\\\\ = [77x -21\dfrac{x^2}{2}]_{0}^{1}\\\\=77-\dfrac{21}{2}\\\\=66.5\)

Thus Volume = 66.5 cu.units.

Find the volume of the solid that lies under the plane 3x + 2y + z = 12, 3x + 2y + z = 12 and above the rectangle R = {(x, y) |0 ≤ x ≤1, - 3 ≤ y ≤ 4}

Summary:

The volume of the solid that lies under the plane 3x + 2y + z = 12, 3x + 2y + z = 12 and above the rectangle R = {(x, y) |0 ≤ x ≤1, - 3 ≤ y ≤ 4} is 66.5 cu.units..