DOUBLE INTEGRALS MATH 200 GOALS Be able to compute double integral - - PowerPoint PPT Presentation

DOUBLE INTEGRALS

MATH 200 WEEK 7 - FRIDAY

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.

MATH 200

INTEGRATION IN CALC II

lim

n→∞ n

• k=1

f(x∗

k)∆x =

b

a

f(x) dx = F(b) − F(a)

RIEMANN SUM FUNDAMENTAL THEOREM OF CALCULUS

MATH 200

DOUBLE INTEGRALS

▸ Rather than integrating over

intervals on the x-axis (or y- axis), we will be integrating

• ver 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

(xk,yk)

Area of each is ΔxΔy Region: R Surface: f(x,y)

EACH PRODUCT IS THE VOLUME OF A RECTANGULAR PRISM

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)

MATH 200

AN EXAMPLE

▸ 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:

Region: R Surface: f(x,y) This is the (net-signed) volume we’re finding

3

1

1 (x2 + y2) dx dy

MATH 200

OKAY, NOW HOW DO WE ACTUALLY INTEGRATE THAT

▸ This integral can be viewed as one integral nested in

another:

3

1

1 (x2 + y2) dx dy =

⇒ 3

1

1 (x2 + y2) dx

• dy

▸ 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: 3

1

1 (x2 + y2) dx

• dy =

3

1

• x3

3 + xy2

• 1
• dy

= 3

1

1 3 + y2 − (0 + 0)

• dy

= 3

1

1 3 + y2

• dy

MATH 200

▸ Now we just have a Calc II integral to evaluate: 3

1

1 3 + y2

• dy = 1

3y + y3 3

• 3

1

= (1 + 9) − 1 3 + 1 3

• = 28

3 ▸ 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…

1 3

1

(x2 + y2) dy dx

MATH 200

1 3

1

(x2 + y2) dy dx = 1

• x2y + y3

3

• 3

1

• dx

= 1

• 3x2 + 9 −
• x2 + 1

3

• dx

= 1

• 2x2 + 26

3

• dx

= 2x3 3 + 26 3 x

• 1

= 2 3 + 26 3

• − (0 + 0)

= 28 3

MATH 200

LET’S TRY ANOTHER ONE

π 2

1

x sin y dx dy = π x2 2 sin y

• 2

1

dy = π 2 sin y − 1 2 sin y dy = 3 2 π sin y dy = −3 2 cos y

• π

= −3 2(−1 − 1) = 3

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

• n the right and left by

two curves x1(y) and x2(y) from y=c and y=d

MATH 200

TYPE 1 REGIONS

▸ Order of integration: dydx

• R

f(x, y) dA = b

a

y2(x)

y1(x)

f(x, y) dydx

x=a x=b y1(x) y2(x)

MATH 200

TYPE 2 REGIONS

▸ Order of integration: dxdy

• R

f(x, y) dA = d

c

x2(y)

x1(y)

f(x, y) dxdy

y=c y=d x1(y) x2(y)

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

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

• R

f(x, y) dA = 1 x

x2 xy2 dydx

1 x

x2 xy2 dydx =

1

• 1

3xy3

• y=x

y=x2

• dx

= 1 1 3x

• (x)3 − (x2)3

dx = 1 1 3x

• x3 − x6

dx = 1 3 1

• x4 − x7

dx = 1 3 1 5x5 − 1 8x8

• 1

= 1 3 3 40

• = 1

40

MATH 200

TAKE ANTIDERIVATIVE WITH RESPECT TO Y PLUG IN Y=X AND Y=X2 AND SUBTRACT SIMPLIFY INTEGRATE WITH RESPECT TO X

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

• R

f(x, y) dA = 1 y1/2

y

xy2 dxdy

1 y1/2

y

xy2 dydx = 1

• 1

2y2x2

• x=y1/2

x=y

• dx

= 1

• 1

2y2 (y1/2)2 − (y)2

• x=y1/2

x=y

• dx

= 1 1 2y2 y − y2 dx = 1 2 1

• y3 − y4

dx = 1 2 1 4y4 − 1 5y5

• 1

= 1 2 1 20

• = 1

40

MATH 200

TAKE ANTIDERIVATIVE WITH RESPECT TO X PLUG IN X- BOUNDS AND SUBTRACT SIMPLIFY INTEGRATE WITH RESPECT TO Y

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

REGION OVER WHICH WE’RE INTEGRATING THE VOLUME WE’RE FINDING WHEN WE INTEGRATE

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

MATH 200

▸ Setup as Type 1 region:

• R

f(x, y) dA = 1/3 3x

x/2

xy2 dydx + 2

1/3

1

x/2

xy2 dydx ▸ How about setting this up as a Type 2 region?

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

• R

f(x, y) dA = 1 2y

y/3

xy2 dxdy

MATH 200

1 2y

y/3

xy2 dxdy = 1 1 2y2x2

• 2y

y/3

dy = 1 1 2y2

• (2y)2 −

y 3 2 dy = 1 2 1 35 9 y4dy = 35 18 1 5y5

• 1

= 7 18(1 − 0) = 7 18

MATH 200

A TRICKIER EXAMPLE

▸ Consider the double integral √π/2 √π

2y

sin(x2) dxdy

▸ 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

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 2 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)

MATH 200

▸ Now the question is, “Can we integrate this as a dydx

integral?”

√π x/2 sin(x2) dydx = √π sin(x2)y

• y=x/2

y=0

dydx = √π sin(x2) x 2 − 0

• dydx

= 1 2 √π x sin(x2)dydx ▸ We can use u-substitution! u = x2; du = 2x dx x = 0 = ⇒ u = 0; x = √π = ⇒ u = π;

MATH 200

√π x/2 sin(x2) dydx = √π sin(x2)y

• y=x/2

y=0

dx = √π sin(x2) x 2 − 0

• dx

= 1 2 √π x sin(x2)dx = 1 2 · 1 2 π sin(u) du = −1 4 cos(u)

• π

= −1 4(−1 − 1) = 1 2

MATH 200

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.

MATH 200

▸ Setting this up as a Type 1 integral, we get

1 4−x2

3x

1 dydx = 1 y

• y=4−x2

y=3x

dx = 1

• (4 − x2) − (3x)
• dx

= 4x − 1 3x3 − 3 2x2

• 1

= 4 − 1 3 − 3 2 = 13 6

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… 1

• (4 − x2) − (3x)
• dx
• R

1 dA = Area of R

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