MA1132 (Advanced Calculus) Solutions to tutorial 10
[April 2–3, 2013]
1.
(a) Express the volume of the ellipsoid
x2 y 2 z 2
+ 2 + 2 ≤1
a2
b
c
as an iterated integral (i.e. set up the triple integral in terms of single
RRRintegrals, but do
not evaluate it). [Hint: The volume of a (solid) region D ⊂ R3 is
1 dx dy dz]
D
Solution: We integrate first with respect to z, keeping (x, y) fixed. The limits for z
arise from expressing the equation for the surface of the ellipsoid in the form
r
x2 y 2
z = ±c 1 − 2 − 2
a
b
Having integrated with respect to z, we must then take the double integral of the
result over the vertical projection (or “shadow”) of the ellipsoid onto the xy-plane.
This projection is, in this case, the section of the ellipsoid through z = 0, or the inside
of the ellipse
x2 y 2
+ 2 = 1.
a2
b
In this way we arrive at the answer
Z x=a Z y=b√1−x2 /a2 Z
√
x=−a
y=−b
1−x2 /a2
√
z=c
!
1−x2 /a2 −y 2 /b2
√
z=−c
1 dz
!
dy
dx.
1−x2 /a2 −y 2 /b2
We could equally do the integrals in another order. If we did the integral dx first, then
dy and finally dz we would get
√ 2 2 2 2
√ 2 2 Z
! !
Z
Z
z=c
y=b
1−z /c
z=−c
y=−b
√
1−z 2 /c2
1−y /b −z /c
x=a
√
x=−a
1 dx
dy
dz.
1−y 2 /b2 −z 2 /c2
Regardless of the order of integration, we obtain integrals that are painful to evaluate
in cartesian coordinates.
(b) Repeat this for the positive ‘octant’ of the same ellipsiod, that is with the added
conditions x ≥ 0, y ≥ 0 and z ≥ 0.
Solution:
√ 2 2 2 2
√ 2 2 Z
! !
Z
Z
x=a
y=b
1−x /a
z=c
1−x /a −y /b
1 dz
x=0
y=0
z=0
dy
dx.
2. Find the mass of a solid object occupying the region in space bounded by the coordinate
planes and the plane x + y + z = 2 if its density function is ρ(x, y, z) = x2 . [Hint: The
mass is the triple integral of the density over the region. Calculations are easier if you leave
the dx integral to last.]
Solution: We know that the answer is
ZZZ
ZZZ
Mass =
ρ(x, y, z)dV =
ρ(x, y, z) dx dy dz
with the triple integral extending over the object.
Leaving the integration with respect to x last means that the first integral to be written is
with respect to x, i..e we have to find the bounds for x. Note that the plane x + y + z = 2
intersects the Ox axis (where y = z = 0) at x = 2, so x goes from 0 to 2. For fixed x the
triangular region to which y and z belong has the boundary given by the lines y = 0, z = 0
and z = 2 − x − y. Thus, for fixed x, the range of values for y is from 0 to 2 − x, and once
x and y are fixed, the range of values for z is from 0 to 2 − x − y.
That gives
Z
2
Z
2−x
Z
z=2−x−y
2
x dz
dy dx
Z
2 z=2−x−y
=
x z z=0
dx dy
x=0
y=0
Z 2 Z 2−x
2
x (2 − x − y) dy dx
=
x=0
y=0
z=0
2 Z 2−z x=0
2
Z
=
Zx=0
2
=
=
=
=
=
y=0
2
2−x
x ((2 − x)y − y 2 /2) y=0 dx
x2 ((2 − x)2 − (2 − x)2 /2) − 0 dx
Zx=0
Z
2
1 2 2
2
2
x (2 − x) /2 dx =
x (4 − 4x + x2 ) dx
2
x=0
x=0
Z 2
1
4x2 − 4x3 + x4 dx
2 x=0
2
1
(4/3)x3 − x4 + x5 /5 x=0
2
1 32
32
8
− 16 +
−0 =
2 3
5
15
2