A contour map is shown for a function f
on the square region R.
notice how we have this closed interval
by this closed interval, where this first interval
would be the interval over x
and this would be the interval for y.
So in our case, notice how the intervals are the same.
So we have a square region R again from zero to four
along the x axis and zero to four along the y axis
gives us the square region,
which in our case would be the region of integration.
We're asked to use the Midpoint Rule
with m equals n equals four to estimate the value
of the double integral over the region R of f of x comma y.
The Midpoint Rule is shown here
where we approximate this double integral by determining
this double sum where we have f of x sub I bar comma
y sub j bar times delta A,
where each product here represents the volume
of a rectangular prism or cuboid
as long as f of x comma y is above the x y plane.
But here because m equals four and so does n,
because m equals four, we divide this horizontal interval
into four equal partitions.
So we'll cut it in half and then half again.
And then because n equals for, we do the same
for the interval for y from zero to four.
So we'll divide it here and here.
So because m equals n equals four,
notice how this divides the region R into 16 smaller regions
or 16 smaller partitions.
So looking at our Midpoint Rule,
delta A would be the area of each smaller partition.
Notice how each partition is a one-by-one square.
So we know delta A int this case is equal to one.
And then x sub i bar comma y sub j bar
would be the midpoint in each partition,
which would be here, here, here, and so on.
So we're going to be using these coordinates
to determine the function value,
which we'll use when approximating the double integral
using the Midpoint Rule.
And luckily notice how each of these midpoints
does fall on a contour line or a contour curve.
So we know the exact function value at that location.
To get a better idea of how the Midpoint Rule works,
let's look at an animation.
Let's the function f of x comma y is this blue surface
pictured here.
Notice how because it is non-negative
or it's above the x y plane,
we could approximate the double integral of f of x comma y
over this square region here using the Midpoint Rule
by partitioning the region of integration as we see here
into 16 smaller partitions.
And then to approximate the value of the double integral,
we would determine the function values
at the midpoint of each square
and then multiply that by the area of each partition,
which gives us the volume of each
of these rectangular prisms or cuboids we see here.
So we use the volume of the cuboids
to approximate the volume under the surface
over the same region.
We can still use the Midpoint Rule
to approximate our double integral,
even if the function f of x comma y is not
above the x y plane, but it would not be interpreted
as volume.
So going back to our example.
Let's set this up on the next slide.
The double integral over the region R of f of x comma y
is going to be approximately equal to,
we have a double sum where j is equal to one to four
and i also equals one to four,
of f of x sub I bar comma,
y sub j bar times delta A.
Again delta A is the area of each smaller partition
which we already know is equal to one.
Let's write this out and write the coordinates
for each of these 16 points.
So the double integral is going to be
approximately equal to, this point here
would have coordinates one half comma one half.
So we'd have f of one half comma one half
times delta A which is one, plus f of,
notice how here j is the inner index.
So the next point would be, x sub one bar
comma y sub two bar, which would be this point here.
So it's f of one half comma three halves
times delta A, which is one, plus f of,
next would be this point here.
f of one half comma five halves times one
plus f of one half comma, this point here
would have a y coordinate of seven halves
times delta A plus next,
i takes on the value of two.
So we have x sub two bar comma y sub one bar,
which would be this point here.
So we have f of three halves comma one half
times delta A plus, next would be,
this point working our way upward.
So we have f of three halves comma three halves
times one plus f of,
again x stays at three halves, y increases to five halves
times delta A which is one,
plus f of three halves comma seven halves times one
plus, now the value of i changes to three.
So we have f of x sub 3 bar comma y sub one bar,
which would be this point here,
and work our way up this column.
So f of five halves comma one half times one
plus f of five halves comma three halves
times one plus f of five halves comma five halves
times one plus f of five halves comma seven halves
times one, hopefully you get the idea here,
plus, finally i takes on the value of four.
So we have f of x sub four bar comma y sub one bar,
which would be this point here.
So f of, this would be seven halves comma one half
times one and so on.
Now we'll determine the function value
using the contour plot, or level curves.
So for f of one half comma one half,
notice how the function value would be 11.
So 11 times one is just 11.
And then from here we have f of one half comma three halves,
which would be here a function value of four
plus f of one half comma five halves here, which is 19
notice f of one half comma seven halves is also 19.
Plus, now we're going to start here.
This is going to be f of three halves comma one half.
Working our way up we'll have 19 plus 11 plus four
plus four.
Next we're here at f of five halves comma one half.
So the function value is 19.
Then we have 11, four, and four.
And then finally we're here at f of seven halves
comma one half, which is 11.
And then we have four, 19, and 19.
This sum comes out to 182,
which is the approximate value of the given double integral
over this region.
I hope you found this helpful.