Setting up an Integral Over a Solid with Order of Integration dz dx dy

Example

Let T be the solid formed by taking all nonnegative (x,y,z) under the plane 3(x-2) + 2y + 6z = 0. It is the tetrahedron with vertices (0,0,0), (2,0,0), (0,3,0), and (0,0,1).

tetrahedron with vertices (0,0,0), (2,0,0), (0,3,0), and (0,0,1)

Basic Steps with Order dz dx dy

To determine the limits of integration, we take the outermost variable and work inward. The integral will have the general form

integral from a to b, g(y) to h(y), j(x,y) to k(x,y) of f(x,y,z) dz dx dy
  1. Determine the maximum and minimum values of the outermost variable. These will be the limits of integration on the first integral sign.

    triple integral over T of f(x,y,z) dz dx dy with arrows to dy and outer integral sign

    Since y is between 0 and 3, the limits of the first integral sign are 0 and 3.

    integral from 0 to 3, g(y) to h(y), j(x,y) to k(x,y) of f(x,y,z) dz dx dy

  2. View a slice formed by keeping the outermost variable constant. Now determine the maximum and minimum values of the middle variable within that slice in terms of the outermost variable. This will give the limits of integration for the middle integral. Note that if the maximum and minimum values depend on where the slice is taken, you will need to split the integral.

    triple integral over T of f(x,y,z) dz dx dy with arrows to dx and middle integral sign

    The slice formed by keeping y constant is a triangle.

    tetrahedron with y slice

    The line formed where the plane 3(x-2) + 2y + 6z = 0 intersects the xy plane is x = -2y/3 + 2. This is found by letting z = 0 in the equation 3(x-2) + 2y + 6z = 0.

    tetrahedron with intersection of plane with xy plane labeled x = -2y/3 + 2

    Therefore we can label the bottom vertex of the triangle formed by the slice.

    y slice, 2 dimensional view for maximum and minimum values of x

    Since x is between 0 and -2y/3 + 2, the limits of the middle integral sign are 0 and -2y/3 + 2.

    integral from 0 to 3, 0 to -2y/3 + 2, j(x,y) to k(x,y) of f(x,y,z) dz dx dy

  3. Finally, using the same slice, determine the range of the innermost variable in terms of the other two variables. This will give the limits of integration for the inner integral. Note that if the range of the innermost variable changes within the slice, you will need to split the integral.

    triple integral over T of f(x,y,z) dz dx dy with arrows to dz and inner integral sign

    The hypotenuse of the triangle formed by taking a slice of the solid with y constant lies on the plane 3(x-2) + 2y + 6z = 0. We solve this equation for z to get z = -x/2 - y/3 + 1.

    y slice with hypotenuse labeled z = -x/2 - y/3 + 1

    Since z is between 0 and -x/2 - y/3 + 1, these are the limits of the innermost integral sign.

    y slice, 2 dimensional view for range of z

    Thus the final integral is as follows:

    integral from 0 to 3, 0 to -2y/3 + 2, 0 to -x/2 - y/3 + 1 of f(x,y,z) dz dx dy

Other Orders of Integration

Back to Integrating Using Rectangular Coordinates.
Back to Describing Surfaces Using Different Coordinate Systems.