Welcome to a lesson on switching the order of integration for double integrals.
In this video we will redo the problem solve in the video entitled
double integrals and volume over a general region
part one.
So you may also want to watch that video to compare the two different methods.
So for this first example.
Our function is z equals x plus y.
And the region is defined on the close interval for x from zero to y over two.
And from y on the close interval from zero to six.
To go back to the previous presentation.
Here's the same problem where we solved it
integrated with respect to x first and then y.
And the answer was 45 cubic units.
So we already know the answer should be 45.
But now I want to integrate this function with respect to y first and then x.
Let's start by taking a close look at this interval here.
This common inequality can be written
so that x is
greater than or equal to zero and
x is less than or equal to y over two.
So we know if x is positive we're gonna be in the first quadrant.
Let's go ahead and solve this inequality for y.
We multiply both sides by two we would have two x.
Is less than or equal to y.
Or if we want it to we can write this as
y is greater than or equal two x.
So y greater or equal to two x would be this blue line here.
And we would shade above the line because y is greater than or equal to two x.
And since y is the interval from zero to six this is our region of integration as we saw before.
This line here is key to setting up this double integral correctly.
This line can be expressed as y equals two x.
Or
x equals y over two.
This would be x equal zero.
Along here.
This point here would be three, six.
And this segment here would be y equals six.
So let's see if we can set this up
integrating respect to y first and then x.
Function is x plus y.
If we would like to integrate with respect to y first.
We must write the limits of integration in terms of x.
So looking at how this region is bounded vertically.
It's bounded below
by
either y equals two x or x equals y over two.
But does it has to be express in terms of x were gonna use two x here.
And then above its bounded by y equals six.
And then for x the integral is from zero to three.
And now we know this should equal 45 but let's go ahead and check it.
Treating x as a constant we'll have xy
plus y squared over two.
Now were gonna replace y
with
six and two x.
So were gonna have six x.
Plus
six squared over two that's gonna be 18.
Minus
Now we'll replace y with two x we'll have two x squared.
And here we're gonna have four x squared divided by two or two x squared.
Let's go and combined our like terms and then we'll integrate with respect to x.
Looks like we'll have negative four x squared.
Plus six x plus 18.
Let's go and complete this on the next slide.
So we'll have negative four times x cubed over three.
Plus six times x squared over two plus 18 x.
Evaluate this at three and zero.
Let's go and rewrite this we have negative four third x cubed.
Plus three x squared.
Plus 18 x.
Replacing x with three we'll have negative four third times
three to the third.
Plus three times
three squared plus 18 times three.
Minus and then when x is zero all these terms will be zero.
This is gonna be negative 36.
Plus
27
plus
54
and this does equal 45.
By switching the order of integration
we did get the same answer.
But what we had to do is make sure that we express the limits of integration for y.
In terms of x. So we had, so we did had to manipulate the equation
x equals y over two.
Let's go ahead and try another one.
Here we want to determine the volume under this surface defined by this region here.
And we want to integrate respect to x first and then y.
Let's go back to our previous presentation to see what the answer is.
Integrate with respect to y and then x.
Again here's the same problem where we had dy dx.
And the answer came out to one twenty fourth cubic unit.
Again we used dy dx
and now I want the order to be dx dy.
Our region of integration is bounded by
the functions
y equals x squared and y equals x.
So here is y equals x.
And here's
y equals x squared.
And the point of intersection here was the point one, one.
So here's our region of integration.
Let's see if we can set this up using the given
order of integration.
Our function is xy.
And then were gonna use dx dy.
So limit of integration
for x. Need to be written in terms of y.
Which means we have to solve these equations for x.
Well this equation is already solve for x we have x equals y.
And then here to solve for x we have to square root both sides of the equation.
We would have x equals the square root of y.
So lower limit of integration for x is going to be
y.
And the upper limit of integration
from this curve here.
Is going to be the square root of y.
Member we have dx here. We have to have these written in terms of y.
And then for the limits of integration for y.
We just have from zero up to one.
And again we know from the previous presentation this is gonna be equal to one twenty fourth.
If we do it correctly.
Let's go and give it a try.
So treating y as a constant.
Were gonna have x squared y over two.
And since we integrate with respect to x were wanna replace x with the square root of y and y.
Replacing x with the square root of y we're gonna have
y squared over two.
Minus here we have y to the third over two.
So were gonna have one half
times y to the third over three.
Minus one half times y to the fourth.
Over four.
Evaluated at one and zero.
Let's finish this on the next slide.
So here we have one sixth y to the third.
Minus one eighth
y to the fourth.
So when y is one we'll have one sixth minus one eighth.
Minus and then when y is zero these are both be zero.
So we have one sixth minus one eighth.
Common denominator of 24.
So we have four twenty fourth minus three twenty fourth.
So it looks good we have one twenty fourth
just as before.
If we take a look at initial problem again.
Since y was already expressed in terms of x.
As we saw in the previous video.
The order of dy dx would of been a little bit less work.
But if were forced to
integrated in this order.
We have to make sure that we expressed x
in terms of y.
Because
because if we have dx here.
Limit of integration must be expressed in terms of y.
And we saw this in the previous example as well.
I hope you found these examples helpful.