| Understanding |
|---|
represents the height of the rectangular solid and
represents the base area. Thus
.
|
Given a closed bounded region
and continuous functions
on
, and suppose that
on
. Then the volume of solid bounded by the lines parallel to
-axis through the boundary
and the surfaces
is given by
.
SOLUTION
Projection of the solid bounded by two cylinders onto
-plane is
and
. Thus,
. Also
implies
. Thus the upper surface is
and the lower surface is
over
. From this, the volume of the solid is set up by
, where
represents the small rectangle of base area and
represents the height of a small solid cylinder.
Now use vertically simple region for
. Then
![]() |
![]() |
![]() |
|
![]() |
![]() |
||
![]() |
![]() |
||
![]() |
![]() |
Take a point in
and evaluate the value of
to find out which surfaces upper or lower surface.
and
.
| Exercise5-8 |
|---|
|
SOLUTION Note that the solid is bounded by the surface goes through the boundary of
and parallel to the
-axis, and the plane
,
.
Using the polar coordinate to express
. Since
,
![]() |
![]() |
![]() |
|
![]() |
![]() |
||
![]() |
![]() |
The limit of integration is constant, we can express the double integral as the product of two single integral.
(a) The region enclosed by
(b) Inside of the curve
and outside of the curve
.
(c) Inside of the curve
and outside of the curve
(a)
with
,
.
(b)
with
.
(c)
with
.
(a)
with
(b)
with
(a) Sphere
with the radius
.
(b)
with
(c)
and
.
(d) The surface generated by rotating
about
-axis for
.
(a)
with
(b)
.
(c)
and
.
(d)
and
,
.