Welcome to a video on setting up
a double integral using both orders of integration.
This video will only focus on
setting up the double integrals.
It will not actually evaluate them.
Here we want to set up the double integrals
using both orders of integration
for the given function over the region
bounded by y equals x squared and y equals x cubed.
So this case, I think it's helpful
to take a look at the graph of y equals x squared
and y equals x cubed, as we see here on the right.
The first thing we need to do is decide
which function is on top and which one is on the bottom.
And in this case, y equals x cubed is on the bottom
and y equals x squared is on top.
And our region of integration is the region
bounded by these two functions.
Now because both of these equations
are already solved for y, or they're functions of x,
it'll be a little bit easier
to determine the double integral
if we integrate with respects to y first.
One more thing before we set this up.
It is important that we determine
this point of intersection here.
Well x squared is equal to x cubed when x is equal to one.
So this point of intersection would be the point one one.
So if we integrate with respects to y first,
we'd have dy here, and then dx.
So our limits of integration for y
must be expressed as functions of x,
and that's why we decided to do this order first.
The lower limit of integration for y would be x cubed
and the upper limit of integration would be x squared.
And then for the limits of integration for x,
we're going to start here at zero and go out to positive one.
So the limits of integration for x will be from zero to one.
Now let's try changing the order of integration.
If we integrate with respects to x first,
we'll be integrating from left to right,
so we'll be integrating from here all the way to here.
Again notice that both boundaries are the two functions,
but now the limits of integration for x
must be a function of y.
So that means we're going to have to take
these two equations here and solve them for x.
So again we can rewrite this as
x equals the square root of y.
And we can rewrite this as x equals the cube root of y.
So our lower limit of integration with respects to x
is going to be the square root of y.
And the upper limit of integration
will be the cube root of y.
And then for the limits of integration for y,
we're starting at zero and going up to positive one.
So one of the most important things to remember here
is if we integrate with respects to y first,
the limits of integration for y must be functions of x.
And if we integrate with respects to x first,
the limits of integration for x must be functions of y.
Let's take a look at another example.
Here's a similar type of problem.
We want to set up the double integrals
using both orders of integration
for the given function over the region bounded
by y equals two x and y equals x squared.
So it is important that we graph these two functions.
I've already graphed it here on the right.
Here we have y equals two x
and here we have y equals x squared.
And then we also want to be able
to find this point of intersection here.
We can look at this graphically
and determine that it's the point two four.
But we could also solve it algebraically
if we set these two equations equal to each other.
We would have two x equals x squared.
Set it equal to zero, factor and solve.
So x equals zero and x equals two.
So when x is zero, we can see that y would be zero.
That's that point here.
And then when x is two, notice both equations
produce a y value of four.
Okay, so here's our region of integration.
Notice it's bounded below and above by these two functions
as well as to the left and to the right.
But since these equations are already
functions of x, or solved for y,
we're going to go ahead and integrate with respects to y first.
So our function is x plus y,
and we're integrate with respects to y first, and then x.
So again if we integrate with respects to y first,
our limits of integration must be functions of x,
meaning y must be solved for x.
And you can see it's already in the correct form.
So our lower limit of integration will be x squared,
and our upper limit of integration will be two x.
And then for the limits of integration for x,
that's why it's important to have this point here.
We'll integrate from zero over to two.
And now let's go ahead and switch the order.
So if we want to integrate now with respects to x first,
our limits of integration must be functions of y
or x expressed in terms of y, which means
we're going to have to solve these two equations for x.
So here if we divide both sides by two,
we'll have x equals y over two.
And here if we solve for x,
we'll square root both sides of the equation.
We'll have x equals the square root of y.
So now we're integrating from left to right.
So again, y equals two x is the same as...
So again y equals two x can be expressed
as x equals y over two.
And y equals x squared can be expressed
as x equals the square root of y.
So when integrating with respects to x first,
we're going to work our way from left to right.
So our lower limit will be y over two,
upper limit of integration with respects to x
will be the square root of y.
And then with respects to y, we'll integrate
from zero all the way to four.
Let's go and take a look at one more example.
This last example is going to be a little bit more challenging.
We want to set up both orders of integration
for the double integral of the given function
over this triangular region.
Let's start by identifying the vertices.
This would be the point zero three.
This would be the point four two.
And this would be the point zero zero.
Now let's go ahead and determine the equation for each side.
This vertical side here would be x equals zero.
This side here is on the line that passes through the origin
and the slope would be up two, right four.
So this would be y equals 1/2 x.
This side here is on the line where the y-intercept is three
and the slope would be down one and right four.
So we'd have y equals negative 1/4 x
plus three.
Notice this region is vertically bounded
only by these two lines,
but horizontally it's bounded by these two lines here,
and then it switches to be bounded by these two lines here.
So let's first integrate this with respects to y
because it's bounded vertically only by two functions.
And it's also nice that these are already functions of x.
So for the lower limit of integration
with respects to y, we'll have 1/2 x.
And the upper limit of integration
will be negative 1/4 x plus three.
And then for the limits of integration with respects to x,
we'll integrate from zero all the way out to positive four.
Now let's go ahead and change the order of integration.
Now if we integrate with respects to x first,
we're working our way from left to right.
Notice below y equals two, the region will be bounded
by this vertical line and this line, y equals 1/2 x.
But above y equals two, it's bounded by this vertical line
and the other line of negative 1/4 x plus three,
which tells us we're going to have to have two double integrals
if we want to integrate with respects to x first.
And we also have to express the limits
of integration for x as functions of y,
which means we'll have to solve these equations for x.
So this one here, we can just multiply
both sides of the equation by two to get x equals two y.
Let's go and take a closer look at this line here.
We'd have to subtract three on both sides
and then multiply both sides by negative four.
So we'll have x equals negative four times y minus three.
So expressed as a function of y,
this would be x equals negative four times y minus three.
Okay, so let's see if we can set this up.
For the first double integral
integrating with respects to x,
the lower limit of integration will be
zero all the way out to two y.
And then for y we'll integrate from zero to two.
And now for this upper region
of the triangle with respects to x,
we'll integrate from zero all the way out
to negative four times the quantity y minus three.
And then for y, we'll integrate from two to three.
That's going to do it for this video.
I hope you found these examples helpful.