Welcome to a lesson
on determining mass using triple integrals.
The mass of an object is the quantity of matter
in an object regardless of its volume
or other forces acting on it.
So, for example, if you leave the planet Earth,
your weight would change, however, your mass would not.
And if an object has uniform density,
then the mass is equal to the density times volume.
However, if a mass has variable density,
we can still determine the mass using triple integrals.
Density given by density function row,
which looks like this P,
then the mass is given by the triple integral
of the density function, integrated over the region D,
where D would represent the boundaries of the solid.
Let's go and take a look at an example.
We want to determine the mass of the solid
in the first octant bounded by the plane
two X plus Y plus Z equals six,
and has the density function row of XYZ
equals zero point five X.
The density is going to be equal to the triple integral
of our density function.
I'm going to go ahead and write this as one half X.
Now, we need to decide on our
order of integration for differential V.
Let's go ahead and use DZ, DY, DX.
We'll start with determining limits of integration for Z.
We know our solid is in the first octant,
and it's bounded by this plane,
So, the lower limit of integration for Z would be zero,
and the upper limit of Z would be determined by this plane,
written as a function of Y and X.
If we solve this equation for Z,
we would have a Z equals six minus two X minus Y,
and that would be our upper limit for Z.
To determine the limits of integration for Y and X,
let's go and take a look at the X Y trace.
Remember, we determined this by setting Z equal to zero,
so we would have the line two X plus Y equals six.
We have an x-intercept of three and a y-intercept of six.
The limits of integration for Y must be
expressed as a function of X,
so we do have to solve this equation for Y.
We would have Y equals six minus two X.
Y starts here at zero, and then it goes up to the line,
again, expressed as a function of X.
The limits of integration for Y,
we would have from zero all the way to six minus two X.
Then, for the limits of integration for X,
the interval would be from zero to three.
The value of this triple integral will give us
the mass of the solid, with density given by this function.
Integrate one half X, with respects to Z,
that would be one half X Z.
Now we'll replace Z with six minus two X minus Y,
and then with zero.
Then, when Z is zero, this would be zero.
Let's continue this on the next page.
Let's go ahead and distribute.
We'll have three X minus X squared,
minus one half X Y.
Now we will integrate with respects to Y.
So, three X Y, minus X squared Y,
and we will have one half X times Y squared over two,
or minus one fourth X Y squared,
and we will replace Y with six minus two X,
and then with zero.
So, three X times six minus two X,
minus X squared times six minus two X.
Here we will have one fourth X
times six minus two X squared.
Let's go ahead and FOIL this here.
We're going to have 36 minus 24 X, plus four X squared.
Then, we're going to multiply this by one fourth X.
We'll have 18 X, minus six X squared,
minus six X squared, plus two X to the third.
Here we are going to multiply this negative one fourth X
times this trinomial, so we will have minus nine X,
this will be plus six X squared,
and in here we are going to have minus X to the third.
Let's go ahead and combine our like terms.
We're going to have one X cubed,
minus six X squared, plus nine X.
Now, we will integrate with respect to X.
We will have one fourth X to the fourth,
minus six times X to the third over three,
that will be minus two X to the third.
Then we are going to have plus nine halves X squared.
Evaluate this at three and zero.
Then when X is zero, we will have zero.
Here we are going to have 81 fourths,
minus 54, plus here, we're going to have
nine times 80 one halves.
This gives us 27 fourths, which is
equal to six point seven five.
Again, this will be the density of the solid.
Again, this would be the density
of the solid in the first octant bounded by this plane here,
with a density function given here.
I hope you found this example helpful.
Thank you for watching.