Use a double integral to find the volume of the solid in the first octant which is enclosed by the surface 3x + 6y + 2z = 12 and the coordinate planes.

8 unit^3

Step-by-step explanation:

Given:

- The equation of the plane is:

                                     3x + 6y + 2z = 12

Find:

Use a double integral to find the volume of the solid in the first octant which is enclosed by the plane and the coordinate planes

Solution:

- Express the equation of surface ( plane ) as a subject of any coordinate axis we will use z:

                                     2z = 12 - 3x - 6y

                                     z = 6 - 1.5x - 3y

- The double integral would be set- up as:

- Where, a , b ,c and d are limits of integration.

- To determine the limits we will project the surface to x-y plane or z = 0 plane, the equation we have is:

                                     0 = 6 - 1.5x - 3y

                                     y = 2 - 0.5x

- For limits a and b the integration is with respect to y, so we express the limits of y in terms of x. Where lower limit b = 0, and upper limit a = 2 - 0.5x

- Similarly, the limits c and d is with respect to x are constants we have:

                                       c = 0

                                       0 = 2 - 0.5*d

                                       d = 4

- Then solve the double integral:

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation: