MATH 253 WORKSHEET 29
TRIPLE INTEGRALS
ex+y+z dV
(3, 0, 0), (0, 2, 0), (0, 0, 1), (0, 0, 0).
xy
plane given by (3, 0, 0), (0, 2, 0), (0, 0, 0) and its apex the point (0, 0, 1). Thus (x, y) will range in
the triangle, and z will range from zero to the value of z on the plane through the points (3, 0, 0),
(0, 2, 0), (0, 0, 1). The equation of that plane is x3 + y2 + z1 = 1 (the coecients of x, y, z chosen so
that the plane passes through the given points). The line connecting (3, 0, 0)and (0, 2, 0) in the xy
y
x
plane is the intersection of this plane with the plane z = 0 so has the equation
3 + 2 = 1. Our
(1) Evaluate
E
Solution:
E
where
is the tetrahedron with vertices
The tetrahedron can be though of as the pyramid with its base the triangle in the
integral is therefore
x=3
y=2(1− x
3)
y
z=1− x
3−2
dy
dx
y=0
x=0
dzex+y+z
x=3
y=2(1− x
3)
dxex
=
z=0
x=0
x=3
y=0
x=0
x=3
x=0
x=3
h
i
y
x
dyey e1− 3 − 2 − 1
y=2(1− x
3)
x
6
dy e1− 3 e 2 − ey
y=0
h
x
1− x
3
y
dxex 2e1− 3 e 2 − ey
=
dxex 2e
x=0
x=3
x=0
x=3
=
=
iy=2(1− x3 )
y=0
x=0
x=3
=
dzez
y=0
dxex
=
y
z=1− x
3−2
z=0
y=2(1− x
3)
dxex
=
dyey
e
1− x
3
− e2(1− 3 ) − 2e1− 3 + 1
x
x
x
x
dxex e2(1− 3 ) − 2e1− 3 + 1
x
2x
dx e2+ 3 − 2e1+ 3 + ex
x=0
ix=3
=
h
=
3e3 − 3e3 + e3 − 3e2 + 3e − 1
x
3e2+ 3 − 3e1+
2x
3
+ ex
x=0
3
= e3 − 3e2 + 3e − 1 = (e − 1) .
Exercise.
(2) Let
E
Do the integral in dierenent orders, for example
be the solid region between the plane
for the integral
E fdV (a) Integrating
Solution:
dy
dz
x=4
z=1
z=0
dz
x=3(1−z)
x=0
and the paraboloid
dx
y=2(1−z− x )
3
y=0
x = y2 + z2 .
dyex+y+z .
Set up the limits
dxf .
x = y2
x-axis.√It
x
2
2
around the x-axis, so the shadow on the xy plane is the disc x + y ≤ 4. Given y, z the x range
starts from a point on the paraboloid and ends at the cap x = 4 at the base of the bullet. The
The region is a solid of revolution: revolve the function
has the shape of a bullet. Slices parallel to the
yz
about the
plane have the shape of discs of radius
integral is therefore
dy dz
y 2 +z 2 ≤4
Date
x=4
dxf =
x=y 2 +z 2
y=+2
dy
y=−2
: 20/11/2013.
1
√
z=+ 4−z 2
√
z=− 4−z 2
x=4
dz
dxf .
x=y 2 +z 2
(b) Integrating
Solution:
at
x
dx dy
dzf .
We now use the observation from before directly: the slice parallel to the
is the disc
y 2 + z 2 ≤ x.
The integral is therefore
x=4
dx
dy dzf =
dx
y 2 +z 2 ≤x
x=0
x=4
√
y=− x
x=0
y=1
z=1−y
dx y=√x dy z=0
x=0
coming from changing the order of integration.
(3) Consider the iterated integral
Solution:
x=1
√
y=+ x
dy
dzf .
yz
plane
√
z=+ x−z 2
√
z=− x−z 2
dzf .
Write the other
We note that the range in the inner integral depends on the
x, y
5
equivalent integrals
chosen before but not
on the order in which they were chosen, so we can switch the rst two variables ignoring the third.
Similarly, xing
x
leaves us with a two-variable integral in which we can switch order as usual. To
start with we switch
graph of
x = y2 ,
y =
√
x
dx dy
to
and below
dy dx.
y = 1,
so if
The domain
x=1
dx
y=1
I2 =
y=1
√
y= x
x=0
√
x ≤ y ≤ 1} is the region above the
y -axis and the graph of
but it is also the region between the
I1 =
we have
{0 ≤ x ≤ 1,
dzf
z=0
x=y 2
dy
z=1−y
dy
z=1−y
dx
y=0
x=0
dzf .
z=0
√
dy√dz in I1√is the integral over the triangle in the yz plane with vertices ( x, 0) , (1, 0)
on the y -axis and ( x, 1 −
x) o it (so the line z = 1 − y passes through√this point and (1, 0)). The
√
z range is thus [0, 1 − x] and the then the y -range is from the line y = x to the line y = 1 − z so
The inner inegral
the integral also equals
x=1
I3 =
√
z=1− x
dz
dx
dyf .
√
y= x
z=0
x=0
y=1−z
We can exchange the rst two variables here: the integral is on the triangular wedge lying below the
√
z = 1 − x in the positive quadrant of the xz plane,
x = (1 − z)2 in the same quadrant. The integral is then
graph of
z=1
I4 =
z=0
x=(1−z)2
dz
dx
y=1−z
√
y= x
x=0
Finally, we can exchange the order of the inner integral in
rectangle
0, y 2 × [0, 1 − z] in the xz
switch the order to get
y=1
and
I4 .
In
I2
the inner integral is on the
so can immediately
x=y 2
dz
z=0
y !)
dxf .
x=0
√
xy plane, above the graph of y = x and
(1 − z)2 , (1 − z) ) (and to the right of the x-axis).
the inner two integrals are in the region of the
below the constant line
Switching the order
y -axis
z=1−y
dy
y=0
I4
I2
dyf .
plane (note the bounds only depend on
I5 =
In
which is also the wedge to the left of
y
y = 1−z
(they meet at
√ 0 to 1 − z
x, so
will range from
and end on the graph of
y=
z=1
I6 =
y=1−z
dz
z=0
and the horizonal lines (slices) will begin at at the
x=y 2
dy
y=0
2
dxf .
x=0