In the following video, we will compute Example 1 and Example 2:
Example 1: Evaluate ∬Rf(x,y)dA,
where f(x,y)=x2y and R is the upper half of the unit
disk (shown here).
Example 2: Evaluate ∬Rf(x,y)dA,
where f(x,y)=4xe2y and the region R is bounded by
the x-axis, the y-axis, the line y=2 and the curve
y=ln(x) (shown here).
Now, we look at some additional examples.
Example 3: Evaluate the integral I∬D(x+y)dA
where D consists of all points (x,y) such that 0≤y≤√9−x2,0≤x≤3.
DO: Graph the region D before reading
further.Then try to set up a Type I iterated
integral.
Solution 3: Since y2=9−x2 is a circle of radius 3 centered at the origin, D
consists of all points in the first quadrant inside this
circle as shown here. This is described as a Type I
region, so we
fix x in and integrate with respect to y along
the black vertical line as shown, and
then integrate with respect to x from x=0 to
x=3 (all possible black vertical lines).
So the double integral I becomes the interated integral ∫30(∫√9−x20(x+y)dy)dx=∫30[xy+12y2]√9−x20dx=∫30(x√9−x2+12(9−x2))dx
DO: Evaluate the above
integral. The answer is 18.
Example 4: Evaluate the integral I=∬R(x+y)dA, where R consists of all points (x,y)
with 0≤y≤3 and 0≤x≤√9−y2.
DO: Graph this region R.
Is it the same as D above? Set up and evaluate the
integral above as a Type II integral over R, which will have
black horizontal lines. You should get the same answer as in
Example 3.