We're asked to consider the function
f of x comma y over the region defined here
which is a triangular region graphed here
in the x, y coordinate plane.
Let's write the equation that will contain
each of these boundaries.
Notice how the equation y equals zero will contain
this side of the triangle.
The equation x equals four will contain this side.
And notice how the line y equals x will contain
this side of the triangle.
This would be on the line with the y-intercept of zero
and a slope of one.
Were asked to evaluate the double integral
over the region D of f of x comma y.
Before we do this though, let's look at the graph
of f of x comma y over this region.
The surface f of x comma y is graphed in blue.
The x, y plane is graphed in yellow.
If we look down on the region of integration,
which is this triangular region here,
notice how we can tell if these surfaces
above the x, y plane over this region,
which means the double integral
is going to be equal to the volume of the solid
below the surface and above the x, y plane
over this triangular region.
So, going back to our work,
let's begin by setting up a double integral.
So the integrand function is y times the square root
of the quantity x cubed plus one.
The given double integral has its integrative respect
to x first then with respect to y.
Because we integrate with respect to x first,
we're integrating from left to right
along the coordinate plane.
So, notice how the left boundary is going to be here
along the line y equals x.
The right boundary is going to be here x equals four.
So, we need to express the limits of integration
with respect to x as functions of y,
so because x equals y, y is the lower limit of integration.
And the upper limit of integration
is going to be x equals four.
And now with respect to y, we're going to integrate
from y equals zero all the way up to this high point,
where y is equal to four.
So, with respect to y, we integrate from zero to four.
If we leave the order of integration in this form,
notice how we need to integrate with respect to x first
treating y as a constant,
but it's going to be quite challenging
to find the antiderivative with respect to x
because we have the square root of the quantity
x cubed plus one.
So for this example, let's go ahead and change
the order of integration.
So the integrand function is going to stay the same.
But let's integrate with respect to y first and then x.
So if we integrate with respect to y first,
notice how the lower boundary's going to be
the line y equals zero and the upper boundary's going to be
this line here where y equals x,
which means the lower limit of integration
is going to be zero.
The upper limit of integration is going to be x
because y is equal to x.
Integrating with respect to x,
notice how the left most point is here
where x equals zero.
So we're going to start integrating at x equals zero
and stop integrating here at x equals four.
So with respect to x we integrate from zero to four.
Now let's go and evaluate this on the next slide.
Integrate with respect to y treating x as a constant,
we're just going to have y squared divided by two
times the square root of the quantity
x to the third plus one.
And I'm going to write the square root
as a rational exponent so we'll have
times the quantity x to the third plus one to the 1/2.
Now these are limits of integration for y,
so we'll substitute x for y then zero for y
and then find the difference.
And let's go ahead and factor out the 1/2,
so we have 1/2 times the quantity,
so when y is equal to x, we have x squared
times the quantity x to the third plus one to the 1/2
minus when y is zero we have zero squared
times the quantity x to the third plus one to the 1/2.
So, of course, this is going to be zero.
So we have 1/2 times the integral from zero to four
of x squared times the quantity x to the third
plus one to the 1/2 dx.
Notice here we do have to perform u substitution,
so we'll let u be equal to x to the third plus one.
So differential u is going to be three x squared dx.
Notice how we have x squared dx.
So let's go ahead and divide both sides by three.
So we have 1/3 du is equal to x squared dx.
So in terms of u, all of this would be again,
x squared dx is equal to 1/3 du.
And because u is equal to x cubed plus one,
this would be u to the 1/2.
So, we'd have 1/2 times with respect to u
we'd have 1/3 times u to the 3/2
and then instead of dividing by 3/2 times 2/3.
So with respect to x, we'd have
1/3 times
u to the 3/2 would really be the quantity x to the third
plus one to the 3/2.
Instead of dividing by 3/2 we'd multiply by 2/3.
Let's go ahead and clean this up a little bit.
1/2 times 1/3 times 2/3 would be equal to 1/9.
So we have 1/9, and here we'd just have the quantity
x to the third plus one to the 3/2,
which we now need to evaluate at four and zero
and then find the difference.
So when x is four, we'd have four cubed plus one to the 3/2
minus when x is zero,
we'd have zero cubed plus one to the 3/2.
So we'd have 1/9 times, here we're going to have,
four to the third is 64, 64 plus one is 65,
so we have 65 to the 3/2 minus,
this would be one to the 3/2 which is just equal to one.
So this would be the exact value.
Let's get a decimal approximation using the calculator.
We have 1/9
times the quantity 65 raised to the power of 3/2,
right arrow, minus one, closed parenthesis,
and ENTER.
Which would be approximately 58.1163.
And of course, we could label this cubic units
because it does represent the volume under the surface
above the x, y plane over the region of integration,
which again would be
the volume of this solid in here.
Okay, I hope you found this helpful.