We're going to use the Midpoint Rule
with m equals n equals two
to estimate the value of the double integral
over the region R of five x plus four y
on the rectangular region R defined here.
When the rectangular region is defined as we see here
using interval notation,
this first interval is the interval for x
and the second interval is the interval for y.
So let's first take a look at our region of integration
and again, x is over the closed interval from zero to four
which would be here
and y is over the closed interval from zero to six
which is here
and therefore, this rectangle is the region of integration.
When using the Midpoint Rule
to approximate a double integral
and our function, f of x comma y is non-negative
over the given region
then we're approximating the volume under the surface
over the given region using cuboids
or right rectangular prisms.
Let's take a look at this graphically.
Our function, f of x comma y is this blue plane
and if we look down on the x, y plane,
we can see the region of integration
is from x equals zero to four
and y equals zero to six.
Because m equals n equals two,
we're dividing both the interval over x and y
into two equal partitions
which give us four sub regions which we see here
or four smaller rectangular regions.
So again, what we're doing when we use the Midpoint Rule
is determining the volume of four right rectangular prisms
or cuboids to approximate the volume
under the surface of the x, y plane
where the smaller partitions in the x, y plane
are the area of the bases
and the height of each cuboid
is determined by the function value
at the midpoint of each partitioned area.
So going back to our work,
let's first find our smaller partitions of the region R.
Again, because m equals two,
we divide the x interval from zero to four
into two equal partitions here
and because n equals two,
we divide the interval from zero to six for y
into two equal partitions which would be here.
Notice how this gives us four smaller partitions
where each of these smaller areas
would be the base of each cuboid
and now we need to find the midpoint
of each of these smaller partitions
which would be here, here,
here and here.
Notice how these points are halfway between
the x values that define the smaller partition
and halfway between the y values
that define the smaller partition.
Let's go ahead and list the coordinates of each midpoint.
Notice how at this point we'd have coordinates,
one comma 1.5.
This point here would have coordinates, one comma 4.5.
This point here would have coordinates, three comma 1.5
and finally, we haven this point here
with coordinates, three comma 4.5.
So looking at the Midpoint Rule defined here,
we have the double sum of f of x sub I bar
comma y sub j bar times delta A
where each of these products represents the volume
of a cuboid or right rectangular prism.
So to set this up, we have the double integral
over the region R of five x plus four y
which would be approximately equal to the double sum
where both j equals one to two
and I equals one to two
of f of x sub I bar
comma y sub j bar times delta A.
Remember, delta A would be the area of the base
of each cuboid which would be the area
of the smaller rectangles here.
So notice how delta A
would be two times three or six
because the area of each of these smaller rectangles
is six square units.
Now, our next step is to find the function values
at each of these points which are the midpoints
of each of these smaller partitions
of the region of integration.
Let's go ahead and do this on the next slide.
So to determine the function values,
our function is f of x comma y equals five x plus four y
so for each point, we substitute the x and y values
into our function to determine the function values.
So we have f of one comma 1.5 is equal to 11,
f of one comma 4.5 equals 23,
f of three comma 1.5 equals 21
and f of three comma 4.5 equals 33.
So using the Midpoint Rule,
we have the double integral over the region R
of five x plus four y
would be approximately equal to,
let's go ahead and write this out as f of
again, one comma 1.5 times delta A
plus f of one comma 4.5
times delta A
plus f of three comma 1.5 times delta A
and then finally plus f of three comma 4.5 times delta A.
So now subbing in our values,
again, f of one comma 1.5 is 11
so we have 11 times delta A which is six
plus f of one comma 4.5 which is 23.
So we have 23 times six
plus f of three comma 1.5 which is 21.
So 21 times six
plus f of three comma 4.5 which is 33.
So we have 33 times six.
Now, we could factor out the delta A here
but I like leaving it in this form
just to emphasize each product
represents the volume of a cuboid
or a right rectangular prism.
So our approximation for the double integral is 528.
So going back to our graph one last time,
the 528 represents the volume of these four cuboids
which we used to approximate the double integral
which represents the volume under the blue plane
over the region of integration.
And again, we can say volume
because our function, f of x comma y,
is non-negative over the region of integration.
Otherwise, we could not refer to it as volume
though we still could use the Midpoint Rule
to approximate a double integral.
I hope you found this helpful.