Last we evaluate the triple integral,
where f of x comma y comma z equals
x plus z divided by three,
u equals the quantity two x minus y divided by two,
v equals y divided by two and w equals z divided by three.
We will use the equations for u, v, and w
to perform a change of variables
in order to evaluate the triple integral.
We'll be using the formula below
to perform the change of variables.
Notice on the left,
we have a triple integral in terms of x, y, and z.
On the right we have a triple integral
in terms of u, v, and w
and also notice how we have the integrating factor
or the Jacobian.
To begin let's list the intervals of integration
for x, y, and z of the original integral.
The interval of integration for x is from y divided by two
to y divided by two plus three.
The interval of integration for y is from zero to three.
And the interval of integration for z
is also from zero to three.
We'll begin by determining the new intervals of integration
or limits of integration for u, v, and w in just a moment.
For the next step,
let's change the form of the equation for u.
Notice right now we have,
u equals the quantity two x minus y divided by two,
which we can write as two x divided by two
minus y divided by two
and therefore u equals x minus y divided by two.
And we'll leave the equations for v and w the same.
I now need to form equations for x, y, and z
in terms of u, v, and w.
We'll notice that y divided by two is equal to v
and therefore we can write the first equation
as u equals x minus v.
And then if we solve this equation for x,
we have x equals u plus v.
And then we'll solve the second equation for y
by multiplying both sides by two,
which gives us y equals 2v.
And then we'll solve the third equation for z
by multiplying both sides by three,
which gives us z equals 3w.
And now we'll begin with
the lower limit of integration for x
or x equals y divided by two
and write this in terms of u.
We now know x equals u plus v,
and we know y divided by two is equal to v.
So solving this equation for u, we have u equals zero,
which gives us a lower limit of integration for u.
And then starting with the upper limit of integration for x,
which is x equals y divided by two plus three,
we substitute u plus v for x.
We substitute v for y divided by two,
and then solve for u, which gives us u equals three.
And therefore three
is the upper limit of integration for u.
Or we can say the interval of integration for u
is from zero to three.
And now we'll find a limits integration for v
using the limits integration for y.
So we know that y is equal to zero.
We also know y is equal to 2v,
which gives us the equation 2v equals zero
and therefore v equals zero.
The upper limit of integration for y is y equals three.
Again, replacing y with 2v and solving
we have v equals 3/2.
So now we know the limits integration for v
or the interval of integration for a v is from zero to 3/2.
And then moving along to z, lower limit for z is equal zero.
We know z's equal to 3w
giving us the equation 3w equals zero.
Solving for w we have w equals zero.
The upper limit is z equals three.
Again, z equals 3w.
Solving for w, we have w equals one.
So now we know the interval of integration
for w is from zero to one.
Next we'll write the function f of x comma y comma z
in terms of u, v, and w and then we'll find the Jacobian.
So we know f of x comma y comma z
equals x plus z divided by three.
We also know x is equal to u plus v
and z divided by three is equal to w,
which gives us a new function of f equals u plus v plus w.
Now for the last step, we need to find the Jacobian.
The Jacobian is equal to the determinant
of the three by three matrix,
where the entries are shown here
in the lower right hand corner.
I've already set this up,
but let's check the entries in the three by three matrix.
In row one, we begin with a partial of x with respect to u,
and we know x is equal to u plus v.
If we differentiate u plus v with respect to u,
treating v as a constant, we do get one.
The next entry is the partial of x with respect to v.
So we differentiate u plus v with respect to v
treating u as a constant, which is also one.
The third entry in row one is the partial of x
with respect to w.
There is no w
and therefore the partial derivative is zero.
In row two we have the partial of y respect to u.
Y equals 2v.
The derivative of 2v with respect to u
treating v as a constant is zero.
Next, we have the partial of y with respect to v.
The derivative of 2v with respect to v is two.
The last entry in row two is the partial
of y with respect to w.
There is no w and therefore the partial derivative is zero.
And then finally in row three,
we have the partial of z with respect to u.
Z equals 3w.
The derivative of 3w with respect to u is zero,
treating w as a constant.
Next, we have the partial of a z with respect to V.
There is no v, the partial derivative is zero.
And then finally we have the partial of z with respect to w.
The derivative of 3w with respect to w is three.
To evaluate the determinant,
I used the expansion by minors method using row two,
since row two had two zero entries.
The three by three determinant is equal to
the entry of two times negative one raised to the power
of two plus two times the determinant
of the two by two matrix, where the entries
are one, zero, zero, three.
Remember we obtain these entries by eliminating the row
and column of the entry two
or eliminating row two and column two,
which leaves the entries of one, zero, zero, three.
So here we have two times negative one to the fourth power,
which is two.
And then times a two by two determinant,
which is equal to three minus zero or three.
Two times three is six.
So now we know that Jacobian
or the integrating factor is six.
Now that we have all the pieces,
we can form the new triple integral and evaluate.
On the left we have the original triple integral.
On the right we have the new triple integral
under the change of variables.
We have the function u plus v plus w times the Jacobian
as the integrand function.
And then we have dudvdw
with limits integration for u from zero to three,
limits of integration for are from zero to 3/2.
The limits of integration for w are from zero to one.
Let's go ahead and distribute the six
before we start integrating.
The integrated function is 6u plus 6v plus 6w
and now we integrate with respect to u
which gives us six times u squared divided by two
or 3u squared plus 6uv plus 6uw.
And now we need to find big F of three
minus big F of w by performing substitution for u.
So when u is equal to three,
we get 27 plus 18v plus 18w.
And when u is equal to zero, we get zero.
And now we integrate with respect to V.
The anti-derivative is 27v
plus 18 times v squared divided by two,
which simplifies to 9v squared plus 18 vw.
We now need to find big F of 3/2 minus big F of zero
by performing substitution for v.
For big F of 3/2 we have 27 times 3/2
plus nine times the square root of 3/2
plus 18 times 3/2 times w which does simplify to 27w.
And big F of zero is zero.
Continuing on the next slide,
if we simplify the integration function,
we get 243/4 plus 27w.
And now I finally integrate with respect to w.
The anti-derivative is 243/4 w
plus 27 times w squared divided by two
or plus 27/2 w squared.
And now we find big F of one minus big F of zero,
which gives us 297/4, or as a decimal 47.25.
I hope you found this helpful.