985
Double Integrals and Volume
EXAMPLE 5
14.2
Volume of a Region Bounded by Two Surfaces
Find the volume of the solid region bounded above by the paraboloid
Paraboloid
z = 1 - x2 - y2
and below by the plane
z = 1- y
as shown in Figure 14.20.
Plane
Paraboloid:
z=1 - x2 - y2
Plane:
z=1-y
Figure 14.20
Solution Equating z-values, you can determine that the intersection of the two
surfaces occurs on the right circular cylinder given by
1- y = 1 - x2 - y2
x² = y - y².
So, the region R in the xy-plane is a circle, as shown in Figure 14.21. Because the volume
of the solid region is the difference between the volume under the paraboloid and the
volume under the plane, you have
Volume = (volume under paraboloid) – (volume under plane)
Wyz?
(1 – x2 - y2) dx dy – (1 – y) dx dy
y-y2
Wy-y²
3
y-y2
R: 0 Sys1
-dy-y2 sxs dy-y2
Figure 14.21
- TT 4
SL
-SI*6 = y2 + x) dx dy
6 – ymla - Kdy
– year2 dy
= 990 1.0 – (23 – 1,276? dy
4
2.
2y - 1 = sin e
(7/2
cos4 e
de
2
-1/2
1/2
cos4 do
1
d."
311
Wallis's Formula
6
16
TT
32
