1. Math 200
Week 7 - Friday
Double Integrals

2. Math 200
Goals
Be able to compute double integral calculations over rectangular regions using partial integration.
Know how to inspect an integral to decide if the order of integration is easier one way (y first, x second) or the other (x first, y second).
Know how to use a double integral as the volume under a surface or find the area or a region in the xy-plane.
Be able to compute double integral calculations over non rectangular regions using partial integration.

3. Fundamental theorem of calculus
Math 200
Integration in calc II
Riemann sum
Fundamental theorem of calculus

4. Each product is the volume of a rectangular prism
Math 200
Double integrals
Rather than integrating over intervals on the x-axis (or y- axis), we will be integrating over regions in the plane.
Given a region R, divide it up into squares with area ΔA = ΔxΔy
Pick a point (xk,yk) in each square
Add the products f(xk,yk)ΔxΔy
Surface: f(x,y)
Each product is the volume of a rectangular prism
(xk,yk)
Area of each is ΔxΔy
Region: R

5. Skipping ahead to definite integrals…
Math 200
Skipping ahead to definite integrals…
Region: R
Surface: f(x,y)
The definite integral of f(x,y) over the region R yields the net-signed volume bounded between R and the surface z=f(x,y)

6. Math 200
An example
Surface: f(x,y)
Consider the paraboloid f(x,y) = x2 + y2 over the rectangular region R = {(x,y) : 0≤x≤1 & 1≤y≤3}
One way to set up a double (iterated) integral for the net- signed volume bounded between R and the surface is like this:
This is the (net-signed) volume we’re finding
Region: R

7. Okay, now how do we actually integrate That
Math 200
Okay, now how do we actually integrate That
This integral can be viewed as one integral nested in another:
For the inner integral, x is our variable, so we’ll take the partial antiderivative with respect to x and evaluate at x=0 and x=1:

8. Now we just have a Calc II integral to evaluate:
Math 200
Now we just have a Calc II integral to evaluate:
So that’s the net-signed volume bounded between f(x,y) and R
We could have set the integral up the opposite way as well…

9. Math 200

10. Math 200
Let’s try another one

11. Non rectangular regions
Math 200
Non rectangular regions
We’ll divide non rectangular regions into two types:
TYPE 1: Regions bounded below and above by curves y1(x) and y2(x) from x=a and x=b
TYPE 2: Regions bounded on the right and left by two curves x1(y) and x2(y) from y=c and y=d

12. Math 200
Type 1 Regions
Order of integration: dydx
x=a
x=b
y1(x)
y2(x)

13. Math 200
Type 2 Regions
Order of integration: dxdy
y=c
y=d
x1(y)
x2(y)

14. Math 200
Example
Say we want to find the volume bounded between f(x,y)=xy2 and the xy-plane over the region bounded by y=x and y=x2

15. Let’s look at the region on the xy-plane:
Math 200
Let’s look at the region on the xy-plane:
We can treat this as a Type 1 region:
It’s bounded below and above by y1(x) = x2 and y2(x) = x
The region extends from x = 0 to x = 1

16. Take antiderivative with respect to y
Math 200
Take antiderivative with respect to y
Plug in y=x and y=x2 and subtract
Simplify
Integrate with respect to x

17. We also could have treated this as a Type 2 region
Math 200
We also could have treated this as a Type 2 region
It’s bounded to the right and to the left x1(y)=y and x2(y)=y1/2
The region extends from y=0 to y=1

18. Take antiderivative with respect to x
Math 200
Take antiderivative with respect to x
Plug in x-bounds and subtract
Simplify
Integrate with respect to y

19. Another example The volume we’re finding when we integrate
Math 200
Another example
Let’s integrate the same function over a different region.
Let f(x,y)=xy2 and take R to be the region bounded by y=3x, y=x/2 and y=1
The volume we’re finding when we integrate
Region over which we’re integrating

20. We want to set the limits of integration to cover the region R
Math 200
We want to set the limits of integration to cover the region R
Notice that we can’t set this up as a single Type 1 region
It’s two Type 1 regions:
(1) y1(x)=x/2; y2(x)=3x
From x=0 to x=1/3
(2) y1(x)=x/2; y2(x)=1
From x=1/3 to x=2

21. How about setting this up as a Type 2 region?
Math 200
Setup as Type 1 region:
How about setting this up as a Type 2 region?

22. The region is bounded on the left by x1(y) = y/3 Solve y=3x for x
Math 200
The region is bounded on the left by x1(y) = y/3
Solve y=3x for x
The region is bounded on the right by x2(y) = 2y
Solve y=x/2 for x
The region extends from y=0 to y=1

23. Math 200

24. A trickier Example Consider the double integral
Math 200
A trickier Example
Consider the double integral
We can’t integrate sin(x2) with respect to x!
But, we can try to switch the order of integration to dydx
Let’s look at what’s going on with the region R:
For the x-bounds we have x1(y) = 2y and x2(y) = sqrt(π)
x = 2y has the same graph as y = x/2
For the y-bounds we have y1 = 0 and y2 = sqrt(π)/2

25. How can we set this up as a Type 1 region?
Math 200
So R is bounded on the left by x1(y) = 2y and on the right by x2(y) = sqrt(π) from y=0 to y=sqrt(π)/2
How can we set this up as a Type 1 region?
R is bounded on the bottom by y1(x) = 0 and on top by y2(x) = x/2
In the x-direction, it extends from x1 = 0 to x2 = sqrt(π)

26. We can use u-substitution!
Math 200
Now the question is, “Can we integrate this as a dydx integral?”
We can use u-substitution!

27. Math 200

28. What happens when we integrate 1
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What happens when we integrate 1
Let f(x,y) = 1 and let R be the region bounded by y=4-x2, y=3x, and x=0 in the first quadrant.

29. Setting this up as a Type 1 integral, we get
Math 200
Setting this up as a Type 1 integral, we get

30. So…? Look at the second line:
Math 200
So…?
Look at the second line:
From Calculus II we know that this integral yields the net- signed area bounded between y=4-x2 and y=3x
So the volume of the solid bounded between f and R is 13/6 units3 BUT it’s also the case that the area bounded between y=4-x2 and y=3x is 13/6 units2
In general…

31. Math 200
Why does this work?
For a cylindrical surface with a base of area A and a height of h, the volume is Ah
In the case that h = 1, V = A
NOTE: V=A numerically, but the units are not the same A h
V = Ah
h=1