Multivariate Calculus - Bruce E. Shapiro

142 LECTURE 17. DOUBLE INTEGRALS OVER GENERAL REGIONS
Solution. This example is exactly like the previous example. Dividing the equation
through by 12 we determine that the given plane intersects the coordinate axes at
x = 4,y = 3, and z = 12. The tetrahedron drops a shadow in the xy plane onto a
triangle formed by the origin and the points (0,3) and (4,0). The equation of the
line between these last two points is
y = −(3/4)x + 3
. Hence the volume of the tetrahedron is given by the integral
V =
∫ 4 ∫ 9−(3/4)x+3
0
0
(12 − 3x − 4y)dydx
Alternatively, we could integrate first over x and then over y as
V =
∫ 3 ∫ −(4/3)y+4
0
0
(12 − 3x − 4y)dxdy
Either integral will give the correct solution. Looking at the first integral,
V =
=
=
=
=
=
=
∫ 4 ∫ −(3/4)x+3
0
∫ 4
0
∫ 4
0
∫ 4
0
∫ 4
0
∫ 4
0
0
(12 − 3x − 4y)dydx
(
12y − 3xy − 2y
2 )∣ ∣ −(3/4)x+2
dx
0
[
12
(− 3 )
4 x + 3 − 3x
(− 3 4 x + 3 )
− 2
(− 3 4 x + 3 ) 2
]
[
−9x + 36 + 9 ( 9
4 x2 − 9x − 2
16 x2 − 9 )]
2 x + 9 dx
[
36 − 18x + 9 4 x2 − 9 ]
8 x2 + 9x − 18 dx
[
18 − 9x + 9 ]
8 x2 + dx
[
18x − 9 2 x2 + 9
(8)(3) x3 ]∣ ∣∣∣
4
= 18(4) − 9 3(64)
(16) +
2 8
= 72 − 72 + 24 = 24
0
Example 17.8 Find the volume of the solid in the first octant bounded by the
paraboloid
z = 9 − x 2 − y 2
and the xy-plane.
Solution The domain of the integral is the intersection of the paraboloid with the
Revised December 6, 2006. Math 250, Fall 2006
dx