We're asked to find the volume of the region
under the surface z equals xy squared
and above the area in the x, y plane bounded
by x equals y squared and x plus y equals six.
So first looking at this graph in the x, y plane,
this curve is x equals y squared
and this line is x plus y equals six.
So the bounded area we're concerned about
is this area here in the x, y plane.
So we're looking for the volume below the surface
above this area here.
Let's look at this in three dimensions
before we set this up.
This graph's a little bit busy,
but the x, y plane is graphed in yellow,
the surface z equals xy squared is graphed in blue.
If you look down on the x, y plane,
the bounded area we're concerned about is this area here.
So we're looking for the volume below the blue surface
above the x, y plane over this area
which we can tell would be the volume in this region here
as well as as this small region here on the left.
So we'll find this volume using a double integral
where the integrand function is xy squared.
Now you need to decide the order of integration.
Looking at the area here, it's going to be easier
to integrate with respect to x first and then y,
because notice horizontally, this area is bounded
by the curve to the left and the line to the right.
So we want to integrate with respect to x first and then y.
If we did integrate with respect to y first,
we'd have to have two separate double integrals
because notice how the area from zero to four
is bounded by this bottom curve and top curve
and then from four to nine the area's bounded
by the line and again the curve.
So it's going to be easier to integrate in this order.
Integrating with respect to x first,
notice how the area's bounded by the curve to the left
and the line to the right.
And because the curve is given by x equals y squared,
y squared is going to be the lower limit of integration
in terms of x.
And for the upper limit of integration,
we need to solve this equation here for x.
Notice solving for x we would have x equals
six minus y which is the upper limit of integration for x.
Now for the limits of integration for y,
notice how this point is the lowest point
with a y coordinate of negative three.
This point is the highest point
with a y coordinate of positive two,
and therefore for y,
we integrate from negative three to positive two.
And now we integrate with respect to x,
treating y as a constant.
So the antiderivative would be x squared divided by two
times y squared or 1/2 x squared y squared.
Remember these are limits of integration for x.
So we substitute these expressions for x
and then find the difference.
So when x equals six minus y, we'd have 1/2
times the quantity six minus y squared times y squared
minus, when x is equal to y squared, we'd have 1/2
times y squared squared times y squared.
Now let's simplify.
Let's go ahead and factor out the 1/2.
So we'd have 1/2 times the integral
from negative three to two of
the quantity six minus y squared
would be 36 minus 12y
plus y squared we still have times y squared,
and then minus we'd have y squared squared
is y to the fourth times y squared
that'd be y to the sixth.
Next we'll go ahead an distribute
and then combine like terms.
And let's do this on the next slide.
So distributing we'd have 36y squared
minus 12y to the third,
plus y to the fourth,
then we have minus y to the sixth.
So we don't have any like terms.
So now we can go ahead and integrate with respect to y.
So we'd have 36 times y to the third divided by three
minus 12 times y to the fourth divided by four
plus y to the fifth divided by five minus y to the seventh
divided by seven.
Let's go ahead and simplify this
before we perform substitution.
So we'd have 1/2 times the quantity here,
we'll have 12y to the third.
This would be minus three y to the fourth
plus, let's write this as 1/5 y to the fifth
minus 1/7 y to the seventh.
So when y is equal to two, we have 12 times two to the third
minus three times two to the fourth,
plus 1/5 times two to the fifth,
minus 1/7 times two to the seventh.
And when y is equal to negative three
we have 12 times negative three to the third,
minus three times negative three to the fourth,
plus 1/5 times negative three to the fifth,
minus 1/7 times negative three to the seventh.
And to save some time I've already determined these values.
This comes out to 1/2 times the quantity,
this simplifies to 1,264/35,
and then we have minus, this comes out to
negative 10,611/35,
and then 1/2 times all of this comes out to
2,375/14, which as a decimal
is approximately 169.6429.
And again, this does give us the volume
bounded by the surface in the x, y plane
over the given bounded area,
which is this volume here as well as this volume
over here on the left.
I hope you found this helpful.