- 1.
- (20 points) The solid
is represented as
the region of space between the graph of
and the
-plane. Find its moment of inertia about the -axis, if
the density is .
Solution: The moment of inertia around the z-axis is the integral
where the shadow region
is
Introducing polar coordinates
this becomes
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- 2.
- (20 points) A solid
has the shape of a
filled ice cream cone: a right circular cone, surmounted by a
slice from a sphere of radius .
The cone has slant height
and a vertex angle
degrees as
pictured.
Take the density to be , and find the gravitational attraction
of
on a unit mass placed at the vertex. (Set up and evaluate
a triple integral in spherical coordinates.)
Solution: By symmetry, the gravitational attraction on a unit mass
at the origin is in the direction of the z-axis and hence equal to its
component. This is
- 3.
- (5 points) Modify the iterated triple
integral of the preceding problem so it now calculates the
gravitational attraction at the vertex due to the part of
lying above the top of the cone (i.e., to the spherical slice).
Just set up the integral; don't evaluate it.
Solution: Instead the integral becomes
- 4.
- (10 points) Suppose the vector field
in
the -plane satisfies
, where
Show that
has no net flux through any simple closed curve .
This should not really have been on this practice exam. Better do one for
flux through a surface.
Solution:
The flux form of Green's
theorem for a simple closed curve
is
Since
by
assumption. Thus the flux integral must vanish.
- 5.
- (45 points; 5, 15, 25)
Let
be the hemisphere of radius , with base in the
-plane and center at the origin, and
be its base (the
disc of radius
and center at the origin in the -plane).
Both surfaces are oriented so their unit normal vectors point
upwards, i.e., this is the direction of positive flux.
Let
.
- (a)
- Find the flux of
over , by inspection.
- (b)
- Using your answer to part (a), find the flux of
over
by using the divergence theorem.
- (c)
- Find the flux of
over
directly, by calculating a
surface integral using spherical coordinates.
Solution:
- (a)
- On
the normal upward normal is
so
and hence the flux upward through
is the area
- (b)
- If
is the upward unit normal to both
and
then
by the divergence theorem
The divergence is
so
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- (c)
- The flux integral written explicitly is
Here
on
so this flux integral is