Find volume of given solid bounded by the coordinate planes and the plane 3x plus 2y plus z equals 6 using double integrals?

Answer 1

To use double integrals to find the volume of a region, you must find a region to integrate over and a surface in terms of two variables to integrate. The surface you are integrating is the plane 3x+2y+z=6. solving this for z to get it as a function of two variables, we get:

z=6-2y-3x

I chose to solve the surface for z because I want the region I am integrating over to be in the xy-plane. This is a purely arbitrary decision, because for this particular problem you could have placed your region in the xz or yz planes, in which case you would solve for the surface in terms of y or x respectively.

The plane in question has x, y, and z axis intercepts all on the positive sides of the axes, and since it is bounded by the coordinate planes, you are examining a solid that is contained within the first octant. We can solve for the axis intercepts of the plane:

By setting x and y equal to zero, we get the z-axis intercept:

z=6-2(0)-3(0); z=6 (0,0,6)

Setting x and z equal to zero gives us the y-axis intercept:

(0)=6-2y-3(0); 2y=6; y=3 (0,3,0)

Setting y and z equal to zero gives us the x-axis intercept:

(0)=6-2(0)-3x; 3x=6; x=2 (2,0,0)

So, in total, the solid we are finding the volume of is a tetrahedron with vertices at (0,0,0), (0,0,6), (0,3,0), and (2,0,0) by examining what exists in the xy-plane of this solid, we can establish a region of integration. If you were to look only at the projection that this shape makes on the xy-plane, you would see a triangle with vertices at (0,0), (0,3), and (2,0).

To define the region, you must choose one variable of the region to define numerically and then the other variable will be dependent on the first variable. I choose to numerically define x. x runs from 0 to 2, as given by the triangle since these are the minimum and maximum values of x within the triangular region. Since x is fixed, we must define y, the other variable in the region, in terms of x. y will run from the x-axis (y=0) to the line segment between (0,3) and (2,0). Using basic point-slope form, we get this line's equation to be:

y-0=(-3/2)(x-2)

y=3-(3/2)x

So, while x runs from 0 to 2, y runs from 0 to 3-(3/2)x

So, the final double iterated integral will read as:

int(0,2)int(0,3-(3/2)x) 6-2y-3x dy dx

This is easily solvable. The antiderivative of 6-2y-3x with respect to y is 6y-y2-3xy, and this must be evaluated from (3-(3/2)x) to 0, so the above iterated integral turns into the single-variable integral:

int(0,2) 6y-y2-3xy|03-(3/2)x dx

int(0,2) [6(3-(3/2)x)-(3-(3/2)x)2-3x(3-(3/2)x)]-[6(0)-(0)2-3x(0)] dx

int(0,2) [(18-9x)-(9-9x+(9/4)x2)-(9x-(9/2)x2)]-[0] dx

int(0,2) [18-9x-9+9x-(9/4)x2-9x+(9/2)x2] dx

int(0,2) [9-9x+(9/4)x2] dx

The antiderivative of 9-9x+(9/4)x2 with respect to x is 9x-(9/2)x2+(3/4)x3

9x-(9/2)x2+(3/4)x3|02

[9(2)-(9/2)(2)2+(3/4)(2)3]-[9(0)-(9/2)(0)2+(3/4)(0)3]

[18-18+6]-[0]

6

So, the final volume of the solid is 6 units cubed.