Welcome to a second example
of changing the order of integration of a double integral.
Noticing its current form,
it would be quite challenging
to integrate with respects to x first
and then with respects to y.
So let's try changing the order of integration.
Let's start by determining the region of integration.
Notice the limits of integration for x
are from y squared to nine.
So that tells us that x
has to be greater than or equal to y squared,
and, it also has to be less than or equal to nine.
And we also know that y
is greater than or equal to zero
and less than or equal to three.
Let's go ahead and graph this region.
So here we have x is greater than or equal to y squared,
let's go ahead and solve this for y.
So we'll square root both sides,
we would have y is less than or equal to
the square root of x.
It would be plus or minus the squared of x
but since we know that y is positive,
the square root of x would also be positive.
So let's go ahead and plot some points.
We'd have the origin, the point 1,1.
When x is four, y is two.
And then when x is nine, y is three.
If we wanted we could switch this around,
y would be less than or equal to the square root of x,
so it'd actually shade below this curve.
And we also know that x is less than or equal to nine.
So here's where x equals nine.
And we would shade to the left of this.
And then for y,
we know that y is greater than or equal to zero,
and then y is less than or equal to three.
Well here's y equals three,
let's go ahead and label this point.
This is the point 9,3.
So this would be our region of integration.
Let's go ahead and label the equations on our graph
y equals the square root of x,
or we could express it as x equals y squared.
Course this was x equals nine,
this was y equals zero.
Now let's go ahead and see if we can switch
the order of integration.
This is x to the 1/2, cosine x.
And now we'll switch it from dxdy to dydx.
So because we're integrating with respects to y first,
our lower limit of integration
will be y equals zero,
and the upper limit of integration would be this curve,
but it must be a function of x,
so we're going to use the square root of x.
Then the limited integration for x
will be from zero all the way to nine.
Now let's see if we can evaluate this double integral.
So the nice thing about this now
is when you integrate with respects to y,
all of this is a constant,
so we're going to have x to the 1/2,
y cosine x,
so that was a much easier anti-derivative to determine.
And now because we integrated with respects to y,
we're going to replace y with the square root of x and zero.
So we're going to replace y with the square root of x,
we're going to have x cosine x,
and then when y is zero, this will be zero.
Let's go ahead and finish this on the next slide.
We are still going to have to use integration by parts,
let's go ahead and review that.
So if we let u equal x,
this should work out pretty well,
let's go ahead and try that.
If u is equal to x,
that means dv
would be equal to cosine xdx,
so differential u would be one dx.
And then v would be the anti-derivative of cosine x,
that's going to be sine x.
So let's see how this works out.
We're going to have u times v,
so we have x sine x,
minus the integral of vdu,
so we'll have sine xdx.
So we're going to have x sine x.
Well, the anti-derivative of sine x
is going to be negative cosine x,
so it'll end up being plus cosine x,
and evaluate this at nine and zero.
So we'll have nine sine nine
plus cosine nine,
minus, we'll have,
zero times sine zero, that's going to be zero,
plus, cosine zero is actually one.
So we have nine sine nine,
plus cosine nine,
minus one.
And that'll do it for this video.