Volume, Mass of an Element(Using Triple Integral)

Presented By:08-EE-09

“If the limits on the integrals involve some of the variables then the order in which the integrations are performed is crucial.”

Some notes on Triple Integrals:

We used a double integral to integrate over a two-dimensional region and so it shouldn’t be too surprising that we’ll use a triple integral to integrate over a three dimensional region.

Since triple integrals can be used to calculate volume, they can be used to calculate total mass (recall Mass = Volume * density) and center of mass

When setting up a triple integral, note that The outside integral limits must be constants The middle integral limits can involve only one variable The inside integral limits can involve two integrals

Example:

A cube has sides of length 4. Let one corner be at the origin and the adjacent corners be on the positive x, y, and z axes.

If the cube’s density is proportional to the distance from the xy-plane, find its mass.

Solution: The density of the cube is f(x,y,z) = kz for some constant k. If W is the cube, the mass is the triple integral.

kkz

kzdz

kzy

kzdydz

dydzkxz

kzdxdydzkzdvw

128|8

16

|4

4

|

40

2

4

0

40

4

0

4

0

4

0

40

4

0

4

0

4

0

4

0

4

0

⇒=

=

=

=

=

=

∫∫

∫∫∫∫

∫∫∫∫∫∫

If distance is in cm and k = 1 gram per cubic cm per cm, then the mass of the cube is 128 grams.

Example:

Let D be the prism bounded by the 3 coordinate planes as well as the planes x = 2 and y + z = 1.

Find the volume of D using triple integrations with all different orders

of x, y and z.

Continued on next slide

Example:

Let D be the solid bounded by the paraboloids z = 8 − x2 − y2, and z = x2 + 3y2. Set up the limits of integrations for evaluating the triple integral:

∫∫∫D

dVzyxF ),,(

Solution:

∫ ∫ ∫∫∫∫=

−=

−=

−−=

−−=

+=

=2

2

2/)4(

2/)4(

8

3

2

2

22

22

),,(),,(x

x

xy

xy

yxz

yxzD

dzdydxzyxFdVzyxF

Example : Evaluate

∫∫∫E

xdv2

Where E is the region under the plane

632 =++ zyx That lies in the first octant. Solution: We should first define octant. Just as the two-dimensional coordinates system can be divided into four quadrants the three-dimensional coordinate system can be divided into eight octants. The first octant is the octant in which all three of the coordinates are positive.

Here is a sketch of the plane in the first octant.

        

We now need to determine the region D in the xy-plane. We can get a visualization of the region by pretending to look straight down on the object from above. What we see will be the region D in the xy-plane. So D will be the triangle with vertices at (0,0),(3,0) and (0,2). Here is a sketch of D.

Now we need the limits of integration. Since we are under the plane and in the first octant (so we’re above the plane z= 0) we have the following limits for z.

                                                         

 

We can integrate the double integral over D using either of the following two sets of inequalities.

yxz 3260 −−≤≤

23

20

30

+−≤≤

≤≤

xy

x

20

32

30

≤≤

+−≤≤

y

yx

Since neither really holds an advantage over the other we’ll use the first one. The integral is then,

9

.1283

4

)3412(

)326(2

2

22

233

0

3

0

23

2

0

22

23

2

0

3

0

326

0

326

0

=

+−=

−−=

−−=

=

=

∫

∫

∫∫

∫∫

∫∫∫∫∫∫

+−

+−

−−

−−

dxxxx

dxxyyxxy

dydxyxx

dAxz

dAxdzxdv

x

x

D

yx

yx

DE

Mass:

The mass of an object can be formulated the same as its volume by introducing the density.

dm = ρdV

Integrating over the distributed mass of the object,

Assuming the density ρ remains constant through out the object we have,

∫∫∫ ρdVm =

m

∫∫∫ dVm = ρ = ρV

V