TRIPLE INTEGRALS MATH 200 GOALS Be able to set up and evaluate - - PowerPoint PPT Presentation

TRIPLE INTEGRALS

MATH 200 WEEK 9 - WEDNESDAY

MATH 200

GOALS

▸ Be able to set up and evaluate triple integrals using

rectangular, cylindrical, and spherical coordinates

MATH 200

TRIPLE INTEGRALS

▸ We integrate functions of

three variables over three dimensional solids

• S

F(x, y, z) dV

dV S (x0,y0,z0)

▸ Chop the solid S up into a

bunch of cubes with volume dV

▸ Pick a point in each cube and

evaluate F there

▸ Add up all of these products

(F•dV)

MATH 200

INTERPRETATIONS

▸ If we think of F(x,y,z) as

giving the density of the solid S at (x,y,z), then the triple integral gives us the mass of S

▸ If F(x,y,z) = 1, then the

integral gives us the volume of S

dV S (x0,y0,z0)

1/3 π 1 zx sin(xy) dzdydx = 1/3 π 1 2z2x sin(xy)

• z=1

z=0

dydx = 1/3 π 1 2x sin(xy) dydx = 1/3 1 2x

• − 1

x cos(xy)

• y=π

y=0

dx = 1/3 −1 2 cos(xy)

• y=π

y=0

dx = 1/3 −1 2 cos(πx) + 1 2 dx = − 1 2π sin(πx) + 1 2x

• 1/3

= − √ 3 4π + 1 6

MATH 200

EXAMPLE 1

d dy cos(xy) = −x sin(xy)

• sin(xy) dy = − 1

x cos(xy) + C

MATH 200

LOTS OF WAYS TO SETUP

▸ Let’s set up a few triple

integrals for the volume of the solid bounded by y2 + z2 = 1 and y = x in the first

• ctant

▸ This means, we’ll just

integrate F(x,y,z) = 1

▸ Here’s what the solid looks

like:

▸ A sketch will really help

with these problems

MATH 200

▸ Let’s say we want to integrate

in the order dzdydx

▸ Once we integrate with

respect to z, z is gone

▸ Visually, we can think of

flattening the solid onto the xy-plane

▸ The top bound for z is the

surface z2=(1-y2)1/2

▸ The bottom bound is the

xy-plane, z1=0

MATH 200

▸ Once we’ve flattened out

in the z-direction, we have a double integral to set up, which we already know how to do!

▸ We have y1=x & y2=1 and

x1=0 & x2=1

▸ So the triple integral

becomes

1 1

x

√

1−y2

1 dzdydx

MATH 200

▸ Alternatively, we could

have gone with dzdxdy

▸ In this case all that

changes is the outer double integral

▸ Going back to the

flattened image on the xy-plane, we get x1=0 & x2=y and y1=0 & y2=1

1 y √

1−y2

1 dzdxdy

yz-plane

MATH 200

▸ We could also not start with

• z. For example, let’s try

dxdzdy

▸ Integrating with respect to x

first will flatten the picture

• nto the yz-plane

▸ On “top” (meaning further

• ut towards us), we have

x2=y

▸ On the “bottom” (meaning

further back) we have x1=0

MATH 200

▸ Now we just set up the

bounds for the outer two integrals based on the flattened image on the yz- plane

▸ z1=0 & z2=(1-y2)1/2 ▸ y1=0 & y2=1 ▸ So the triple integral

becomes

1 √

1−y2

y 1 dxdzdy

MATH 200

▸ Pick one of these three to integrate:

1 1

x

√

1−y2

1 dzdydx 1 y √

1−y2

1 dzdxdy

▸ With the dzdydx integral, we end up needing trig

substitution to perform the second integration, so we should go with the second or third option

1 √

1−y2

y 1 dxdzdy

MATH 200

1 y √

1−y2

1 dzdxdy = 1 y z

• √

1−y2

dxdy = 1 y

• 1 − y2 dxdy

= 1 x

• 1 − y2
• y

dy = 1 y

• 1 − y2 dy

=

1

−1 2 √u du = −1 2 2 3u3/2

• 1

= 1 3

MATH 200

EXAMPLE 2

Evaluate

• G

z√y dV where G is the solid enclosed by z = y, y = x2, y = 4, and z = 0.

MATH 200

▸ Let’s try dzdydx first ▸ z1=0 and z2=y ▸ Flatten the solid onto the xy-

plane

▸ Now for y we have… ▸ y1=x2 & y2=4 ▸ Finally, ▸ x1=-2 & x2=2

2

−2

4

x2

y z√y dzdydx = 2 2 4

x2

y z√y dzdydx = 2 2 4

x2

1 2z2√y

• y

dydx = 2 4

x2 y2√y dydx

= 2 4

x2 y5/2 dydx

= 2 2 7y7/2

• 4

x2 dx

= 2 7 2 128 − x7

• 4

x2 dx

= 2 7

• 128x − 1

8x8

• 2

= 2 7 (256 − 32) = 64

MATH 200

WE HAVE TO BE CAREFUL WHEN USING SYMMETRY: IT WORKS HERE BECAUSE BOTH THE FUNCTION WE’RE INTEGRATING AND THE REGION OVER WHICH WE’RE INTEGRATING ARE SYMMETRIC OVER THE PLANE X=0.

MATH 200

▸ Let’s look at some alternative

setups

▸ We could have started with x

instead and done dxdzdy

▸ If we draw a line through the

solid in the x-direction, it first hits the back half of the parabolic surface and then the front half of the parabolic surface

▸ If we then collapse the picture

• nto the yz-plane, we get this…

4 y √y

−√y

z√y dxdzdy

MATH 200

TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES

▸ Cylindrical coordinates ▸ We already know from polar that dA = rdrdθ ▸ So, for dV we get rdrdθdz ▸ Just replace dxdy or dydx with rdrdθ ▸ Spherical coordinates ▸ For now, let’s just accept that in spherical coordinates, dV

becomes ρ2sinφdρdφdθ

▸ We’ll come back to why this is the case in the next section

MATH 200

EXAMPLE

▸ Consider the integral ▸ First let’s get a sense of what the region/solid looks like ▸ z1 = -(4 - x2 - y2)1/2 and z2 = (4 - x2 - y2)1/2 ▸ Squaring both sides of either equation, we get a sphere of

radius 2 centered at (0,0,0): x2 + y2 + z2 = 4

▸ So we’re going from the bottom half of the sphere to the top

half

▸ y1 = 0 & y2 = (4-x2)1/2 and x1 = 0 & x2 = 2 ▸ On the xy-plane, we go from the line y=0 to the top half of a

circle of radius 2, but only from x=0 to x=2.

2 √

4−x2

√

4−x2−y2 −√ 4−x2−y2 x2 + y2 dzdydx

MATH 200

MATH 200

▸ Setup in cylindrical coordinates ▸ Since z is common to rectangular and cylindrical, let’s

start with that

z = −

• 4 − x2 − y2 =

⇒ z = −

• 4 − r2

z =

• 4 − x2 − y2 =

⇒ z =

• 4 − r2

▸ Now we can look at what remains on the xy-plane and

convert that to polar (recall: cylindrical = polar + z)

▸ y1 = 0 & y2 = (4-x2)1/2 and x1 = 0 & x2 = 2 ▸ On the xy-plane, we go from the line y=0 to the top

half of a circle of radius 2, but only from x=0 to x=2.

▸ r goes from 0 to 2 and θ goes from 0 to π/2

MATH 200

z2 =

• 4 − r2

z1 = −

• 4 − r2

r1 = 0 r2 = 2 θ1 = 0 θ2 = π 2

2 √

4−x2

√

4−x2−y2 −√ 4−x2−y2(x2 + y2) dzdydx =

π/2 2 √

4−r2 − √ 4−r2 r2 rdzdrdθ

MATH 200

▸ For spherical, let’s start with ρ: ▸ The sphere of radius 2 is simply ρ=2 ▸ The region starts at the origin: ρ=0 ▸ Remember, φ measures the angle taken from the positive

z-axis

▸ In order to cover the quarter-sphere, φ needs to go from

0 to π.

▸ We already know what θ does from cylindrical coordinates ▸ The integrand (the function we’re integrating) is a little

more involved…

x2 + y2 = (ρ sin φ cos θ)2 + (ρ sin φ sin θ)2 = ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ = ρ2 sin2 φ(cos2 θ + sin2 θ) = ρ2 sin2 φ

MATH 200

▸ Lastly, we can’t forget about the “extra term,” ρ2sinφ:

2 √

4−x2

√

4−x2−y2 −√ 4−x2−y2(x2 + y2) dzdydx =

π/2 π 2 ρ2 sin2 φ ρ2 sin φ dρdφdθ = π/2 π 2 ρ4 sin3 φ dρdφdθ