Triple Integrals: Setting up the Integral

1. Set up the integral of a function f Hx, y, zL over the region above the upper nappe of the cone z2 = 2 x2 + 2 y2 from z = 2 toz = 6. Use the following orders of integration: d x d y d z, d y d x d z and d z d x d y.

Solution: The region of integration is

ContourPlot3D@8z^2 � 2 x^2 + 2 y^2, z � 2, z � 6<, 8x, -5, 5<, 8y, -5, 5<, 8z, 0, 6.5<,ContourStyle ® 8Directive@Opacity@0.3DD, Directive@Opacity@0.8DD, Directive@Opacity@0.8DD<D

-5

0

5

-5

0

5

0

2

4

6

For the d x d y d z order of integration we have

à2

6

à-z� 2

z� 2

à-

z2-2 y2

2

z2-2 y2

2f Hx, y, zL â x â y â z

To find the y and z limits, we look at the projection of our region onto the yz - plane:

ContourPlot@z^2 � 2 y^2, 8y, -5, 5<, 8z, 2, 6<D

-4 -2 0 2 4

2

3

4

5

6

For a fixed z, we see that y goes from the line on the left to the line on the right. The left line has equation z = 2 y and the line

on the right has equation z = - 2 y. This explains the limits on the second integral. The limits on the z integral are clearly 2 and6.

For the d y d x d z order of integration we have - z2-2 x2

2£ x £

z2-2 x2

2. For the x limits, we have the same picture as that just

above only with the horizontal axis being the x - axis. Hence the integral is

à2

6

à-z� 2

z� 2

à-

z2-2 x2

2

z2-2 x2

2f Hx, y, zL â y â x â z

Finally, for the d z d x d y order of integration, we have a more complicated situation. The integral breaks into two parts:

1. The portion of the region inside the cone that lies above the circle 4 = 2 x2 + 2 y2. Here, z runs from 2 to 6.

2. The portion outside the circle 4 = 2 x2 + 2 y2 and inside the circle 6 = 2 x2 + 2 y2. Here, z runs from the cone up to z = 6.

For (1), the integral is

2 Triple_Integral_Examples.nb

à- 2

2

à- 2-y2

2-y2

à2

6

f Hx, y, zL âz âx ây

since we are directly above the circle of radius 2 centered at the origin and z runs from 2 to 6 over this circle.

For (2) we have, for limits on z, 2 x2 + 2 y2 £ z £ 6. The region in the xy-plane is the annular region shown below:

ContourPlotBx^2 + y^2, 8x, -4, 4<, 8y, -4, 4<,

RegionFunction ® FunctionB8x, y, z<, 2 < x^2 + y^2 < 18 F, BoundaryStyle ® RedF

-4 -2 0 2 4

-4

-2

0

2

4

We can use symmetry to reduce the problem here to the first quadrant and multiply by 4. For any fixed y such that

2 £ y £ 3 2 , x runs from the y-axis to the outer circle: 0 £ x £ 18 - y2 . For any fixed y such that 0 £ y < 2 , x runs from

the inner circle to the outer circle: 2 - y2 £ x £ 18 - y2 . Hence our integral for (2) is:

4 :à0

2

à2-y2

18-y2

à2 x2+2 y2

6

f Hx, y, zL â z â x â y + à2

3 2

à0

18-y2

à2 x2+2 y2

6

f Hx, y, zL â z â x â y>

The final integral for this order of integration is

à- 2

2

à- 2-y2

2-y2

à2

6

f Hx, y, zL â z â x â y +

4 :à0

2

à2-y2

18-y2

à2 x2+2 y2

6

f Hx, y, zL â z â x â y + à2

3 2

à0

18-y2

à2 x2+2 y2

6

f Hx, y, zL â z â x â y>

2. Set up the integral of a function f Hx, y, zL over the region bounded above by z = 8 - x2 - y2 and below by z = -1 + 2 x2 + 3 y2

using the best choice of the order of integration. Find the volume of this region using a triple integral.

Solution: A plot of this region is shown below:

Triple_Integral_Examples.nb 3

2. Set up the integral of a function f Hx, y, zL over the region bounded above by z = 8 - x2 - y2 and below by z = -1 + 2 x2 + 3 y2

using the best choice of the order of integration. Find the volume of this region using a triple integral.

Solution: A plot of this region is shown below:

ContourPlot3D@88 - x^2 - y^2 � z, z � -1 + 2 x^2 + 3 y^2<, 8x, -4, 4<, 8y, -4, 4<,8z, -1.5, 8.5<, Mesh ® None, ContourStyle ® 8Opacity@0.4D, Opacity@0.4D<D

-4

-2

0

2

4

-4

-2

0

2

4

0

5

Looking at this plot, we see that the easiest order of integration is d z d y d x (or d z d x d y). The region in the xy-plane that we

integrate over is found by setting 8 - x2 - y2 = -1 + 2 x2 + 3 y2. We get 9 = 3 x2 + 4 y2 - an ellipse. This ellipse is shown below:

4 Triple_Integral_Examples.nb

ContourPlot@9 � 3 x^2 + 4 y^2, 8x, -2, 2<, 8y, -2, 2<D

-2 -1 0 1 2

-2

-1

0

1

2

Clearly the region in the xy - plane is both x and y simple. Our integral is:

à- 3

3

à-

9-3 x2

4

9-3 x2

4

à-1+2 x2+3 y2

8-x2-y2

f Hx, y, zL â z â y â x

The volume of this region is simply the integral above with f Hx, y, zL = 1. We have:

à- 3

3

à-

9-3 x2

4

9-3 x2

4

à-1+2 x2+3 y2

8-x2-y2

1 â z â y â x

27 3 Π

4

3. Find the volume of the region bounded by the graphs of z = 3 x2, z = 4 - x2, y = 0, and z + y = 6.

Solution: As seen below, the region lies under the cylinder z = 4 - x2, over the cylinder z = 3 x2 , to the right of the xz-plane andto the left of the plane z + y = 6. Hence the order of integration to use here is d y d z d x or d y d x d z.

Triple_Integral_Examples.nb 5

ContourPlot3D@8z � 3 x^2, z � 4 - x^2, z + y � 6, y � 0<, 8x, -1.5, 1.5<, 8y, 0, 6.5<, 8z, 0, 4.5<,ContourStyle ® 8Opacity@0.5D, Opacity@0.5D, Opacity@0.9D, Opacity@0.9D<, Mesh ® NoneD

-1

0

102

4

6

0

1

2

3

4

The region in the xz-plane to integrate over is

Plot@84 - x^2, 3 x^2<, 8x, -1, 1<, AxesLabel ® 8x, z<, Filling ® 81 ® 82<<D

z = 4 - x2

z = 3 x2

-1.0 -0.5 0.5 1.0x

1

2

3

4

z

This plot clearly shows us that we will need to use a d z d x ordering for the outer two integrals. The integral for the volume istherefore

6 Triple_Integral_Examples.nb

à-1

1

à3 x2

4-x2

à0

6-z

1 â y â z â x

The integral evaluates to:

à-1

1

à3 x2

4-x2

à0

6-z

1 ây âz âx

304

15

Triple_Integral_Examples.nb 7