TRIPLE INTEGRALSMATH 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) dVdV

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

0

� �

0

� 1

0zx sin(xy) dzdydx =

� 1/3

0

� �

0

1

2z2x sin(xy)

����z=1

z=0

dydx

=

� 1/3

0

� �

0

1

2x sin(xy) dydx

=

� 1/3

0

1

2x

�� 1

xcos(xy)

�����y=�

y=0

dx

=

� 1/3

0�1

2cos(xy)

����y=�

y=0

dx

=

� 1/3

0�1

2cos(�x) +

1

2dx

= � 1

2�sin(�x) +

1

2x

����1/3

0

= ��

3

4�+

1

6

MATH 200

EXAMPLE 1

d

dycos(xy) = �x sin(xy)

�sin(xy) dy = � 1

xcos(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 octant

▸ 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

0

! 1

x

! √1−y2

01 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

0

! y

0

! √1−y2

01 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 onto the yz-plane

▸ On “top” (meaning further out 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

0

� �1�y2

0

� y

01 dxdzdy

MATH 200

▸ Pick one of these three to integrate:! 1

0

! 1

x

! √1−y2

01 dzdydx

! 1

0

! y

0

! √1−y2

01 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

0

� �1�y2

0

� y

01 dxdzdy

MATH 200� 1

0

� y

0

� �1�y2

01 dzdxdy =

� 1

0

� y

0z

����

�1�y2

0

dxdy

=

� 1

0

� y

0

�1 � y2 dxdy

=

� 1

0x�

1 � y2

����y

0

dy

=

� 1

0y�

1 � y2 dy

=

� 0

1�1

2

�u du

= �1

2

�2

3u3/2

�����0

1

=1

3

MATH 200

EXAMPLE 2Evaluate

���

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

0z�

y dzdydx = 2

� 2

0

� 4

x2

� y

0z�

y dzdydx

= 2

� 2

0

� 4

x2

1

2z2�y

����y

0

dydx

=

� 2

0

� 4

x2

y2�y dydx

=

� 2

0

� 4

x2

y5/2 dydx

=

� 2

0

2

7y7/2

����4

x2

dx

=2

7

� 2

0128 � x7

����4

x2

dx

=2

7

�128x � 1

8x8

�����2

0

=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 onto the yz-plane, we get this…! 4

0

! y

0

! √y

−√yz√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

0

! √4−x2

0

! √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

0

! √4−x2

0

! √4−x2−y2

−√

4−x2−y2

(x2 + y2) dzdydx =

! π/2

0

! 2

0

! √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 = (� sin � cos �)2 + (� sin � sin �)2

= �2 sin2 � cos2 � + �2 sin2 � sin2 �

= �2 sin2 �(cos2 � + sin2 �)

= �2 sin2 �

MATH 200

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

� 2

0

� �4�x2

0

� �4�x2�y2

��

4�x2�y2

(x2 + y2) dzdydx =

� �/2

0

� �

0

� 2

0�2 sin2 � �2 sin � d�d�d�

=

� �/2

0

� �

0

� 2

0�4 sin3 � d�d�d�