Suggested problems
The triple integral
The only realistic way to sketch these is to by-hand sketch, or use a combination of Winplot/MVT and
annotate it a bit. The key features are a good picture of the 2D “floor,” and enough of a 3D picture to
determine what is above what and whether you need to look for points of intersection. Graphs below are
produced in Winplot, and hand annotated as needed.
P1: Evaluate the triple integral
D
6xy dV
where
√ D lies under the plane z = 1 + x + y and above the region in the xy plane bounded by
y = x, y = 0, and x = 1.
For this one, it’s easiest to sketch the floor first and set up the x and y bounds:
√
0≤x≤1
and
0≤y≤ x
The plane z = 1 + x + y is floating above that region (and does not intersect with the xy plane, so
0≤z ≤1+x+y
This sets up the iterated integral
1 √x 1+x+y
0
0
0
6xy dz dy dx
Inner:
1+x+y
0
Middle:
√x
0
Outer:
6xy dz = 6xyz ]01+x+y
= 6xy[(1 + x + y) − 0]
= 6xy + 6x2 y + 6xy 2
√
x
3
(6xy + 6x2 y + 6xy 2 ) dy = 3xy 2√+ 3x2 y 2 + 2xy
]0
√
√
2
2
2
= (3x( x) + 3x ( x) + 2x( x)3 ) − (0)
= 3x2 + 3x3 + 2x5/2
1
0
(3x2 + 3x3 + 2x5/2 ) dx = x3 + 43 x4 + 47 x7/2 ]10
= (1 + 34 + 74 ) − (0)
≈ 2.321
P2: Evaluate the triple integral
D
xz dV
where D is the solid tetrahedron with vertices (0, 0, 0), (1, 1, 0) and (0, 1, 1).
For this one, you need to do a bit of work to determine the equation of the top plane, passing
through the points (0, 0, 0), (0, 1, 0), (1, 1, 0) and (0, 1, 1). This is the vector problem of determining
the equation of a plane through three points - get vectors from (0, 0, 0) to the other two points:
< 1, 1, 0 > and < 0, 1, 1 >. Cross them to get the normal:
< 1, 1, 0 > × < 0, 1, 1 >=< 1, −1, 1 >
and use the point normal:
< 1, −1, 1 > · < x − 0, y − 0, z − 0 >= 0
x−y+z = 0
The, you also will need the equation for the boundary line in the xy plane where that plane crosses:
when z = 0, we have x − y = 0.
The bounds are
0≤x≤1
and the iterated integral is
x≤y≤1
1 1 −x+y
0
x
0
0 ≤ z ≤ −x + y
xz dz dy dx
Inner:
−x+y
0
xz dz = 21 xz 2 ]0−x+y
= 12 x[(−x + y)2 − 02 ]
= 12 x[x2 − 2xy + y 2 ]
= 12 (x3 − 2x2 y + xy 2 )
Middle:
1
x
Outer:
1 (x3 − 2x2 y + xy 2 ) dy = 1 (x3 y − x2 y 2 + 1 xy 3 ) ]1
x
2
2
3
= 12 [(x3 (1) − x2 (1) + 31 x(1)) − (x3 (x) − x2 (x2 ) + 31 x(x3 ))]
= 12 [x3 − x2 + 31 x − x4 + x4 − 31 x4 ]
= 12 [− 31 x4 + x3 − x2 + 13 x]
= − 61 x4 + 12 x3 − 12 x2 + 16 x
1
0
1 x5 + 1 x4 −
(− 61 x4 + 21 x3 − 12 x2 + 16 x) dx = − 30
8
1 +1−1+
= − 30
8 6
≈ 0.0083333
1 3
1 2 1
6 x + 12 x ]0
1
12
P3: Set up (do not evaluate) the iterated integral that will give the the volume of the tetrahedron
enclosed by the coordinate planes and the plane 4x + y + 6z = 10.
10
The intercepts of the tetrahedron are (0, 0, 10
6 ), (0, 10, 0) and ( 4 , 0, 0), and the intersection with
the xy plane (z = 0) is the line 4x + y = 10. This gives a region bounded by
0 ≤ x ≤ 2.5
0 ≤ y ≤ 10 − 4x
0≤z≤
10 − 4x − y
6
The volume of the region is given by
D
dV =
2.5 10−4x (10−4x−y)/6
0
0
0
dz dy dx
P4: Set up (do not evaluate) the iterated integral that will give the the volume of the solid bounded by
the elliptic cylinder 4x2 + z 2 = 4 and the planes y = 0 and y = z + 2.
If you consider the 3d version of this one first, you’ll notice that the cylinder opens along the y axis,
and the best way to visualize it is with the “floor” in the xz plane, and the “top” and “bottom”
being sliced off by the planes y = 0 and y = z + 2.
The region is bounded by
−1 ≤ x ≤ 1
−
√
√
4 − 4x2 ≤ z ≤ 4 − 4x2
0≤y ≤z+2
The volume of the region is given by
D
dV =
1 √4−4x2 z+2
−1
√
− 4−4x2
0
dy dz dx
P5: Set up (do not evaluate) the iterated integrals that will give the the volume of the solid bounded
by the paraboloids z = x2 + y 2 and z = 18 − x2 − y 2 .
You need to find the intersection of the paraboloids:
x2 + y 2 = 18 − x2 − y 2
2x2 + 2y 2 = 18
x2 + y 2 = 9
They intersect in a cicle of radius 3, and this is the region in the xy plane over which you should
integrate.
−3 ≤ x ≤ 3
−
√
9 − x2 ≤ y ≤
√
9 − x2
x2 + y 2 ≤ z ≤ 18 − x2 − y 2
The integral that gives the volume is
D
dV =
3 √9−x2 18−x2 −y2
−3
√
− 9−x2
x2 +y 2
dz dy dx
P6: Write five other iterated integrals that are equal to the given iterated integral:
1 1 y
0
y
0
f (x, y, z) dz dx dy
The bounds of the region can be taken from the integral as written:
0≤y≤1
y≤x≤1
0≤z≤y
And note that since x and z are both functions of only one variable, y, we can swap their order
and still have something that will work out OK:
1 y 1
0
0
y
f (x, y, z) dx dz dy
So two ways to look
at this (consider the outer two variables) are as someting with a floor in the
xy plane (the 01 y1 0y f (x, y, z) dz dx dy version), and something with a floor in the zy plane (the
1 y 1
0 0 y f (x, y, z) dx dz dy version). Switching order of integration on the floor is pretty straightforward:
xy floor (the dx dy, getting switched to a dy dx)
New integral:
0≤y≤1
y≤x≤1
0≤x≤1
0≤y≤x
1 x y
0
0
0
f (x, y, z) dz dy dx
yz floor (the dz dy, getting switched to a dy dz)
0≤y≤1
0≤z≤y
z≤y≤1
0≤z≤1
New integral:
Switch
1 y 1
0
to
0
y
1 1 1
0
z
y
f (x, y, z) dx dz dy
f (x, y, z) dx dy dz
The integrals we are missing are the views that have the xz floor. For that, I need to look at the
3d.
This gives
z≤y≤x
0≤z≤x
1 x x
0
0
z
0≤x≤1
f (x, y, z)dy dz dx
or
z≤y≤x
z≤x≤1
1 1 x
0
z
z
0≤z≤1
f (x, y, z)dy dx dz