Math 2224 - Review for Test 2
I. 1. Sketch the region described by the limits of integration.
9 3
!!
0
y
cos x 3 dxdy
2. Rewrite as an equivalent integral with the order of integration reversed.
II. Change the following to an equivalent integral in polar coordinates.
2
4! y 2
1
"0 "y 1 + x 2 + y 2 dxdy
III. A cylindrical tube of radius 2 cm. and height 3 cm. is spinning around on its axis, causing the gas
inside to “pile up” around the walls of the cylinder. The density of gas at any point in the cylinder is .2
times the distance to the axis of rotation. Find the total mass of the gas in the tube. (Suggestion: Let the
cylinder sit on the xy-plane with the z axis running through it vertically. Use the coordinate system of
your choice to evaluate the integral.)
IV. 1. Under what circumstances is the following true?
b d
! ! f (x, y)
a c
dydx =
d
b
c
a
!!
f (x, y) dxdy
2. Under what circumstances will the following be true?
a
" "
!a
b
a b
f (x, y) dydx = 4"0 "0 f (x, y) dydx
!b
V. Evaluate
1 1 y2
!!!
0 y yz
yz dxdzdy
VI. 1. Set up a double integral to find the volume of the tetrahedron bounded by the coordinate planes
and the plane 2x + 3y + z = 6.
2. Set up a triple integral for the above.
VII. Rewrite
2
" "
!2 0
4! x 2
sin(x 2 + y 2 ) dydx as a polar integral. Evaluate it.
VIII. 1. Sketch the solid in the first octant bounded by the coordinate planes and the graphs of
x + z = 6 and z = 4 - y2.
2. Set up a triple integral in Cartesian coordinates to find the volume.
IX. A solid h as the shape of the cylinder x2 + y2 = 1 cut off at the top by the graph of
x2 + y2 + z2 = 4 and the bottom by the xy plane. The density at any point P(x,y,z) is
δ(x,y,z) = x.
1. Set up a triple integral in cylindrical coordinates to find the mass of the solid.
2. Set up a triple integral in spherical coordinates to find the mass of the solid.
Answers
I. 1.
II.
2.
#
2
4
0 0
1
" " 1+r
3 x2
!!
0 0
cosx 3 dydx
rdrd !
2
III. 3.2 π
IV. 1. All circumstances
1
72
V. !
VI. 1.
VII.
2. If f is symmetric w.r.t. x and y axes
2
3 ! x+2
3
0 0
""
"
2
0
0
6 ! 2x ! 3y dydx
$ $ sin(r )rdrd!
2
=
2.
2.
2$
1
0
0 0
# ##
%
2% 2
6
0 0
0
$ $ $
4"r 2
""
"
1 dzdydx
"
(1 # cos4 )
2
VIII. 1.
IX. 1.
2
2
3 ! x + 2 6! 2x ! 3y
3
0 0
0
2 4! y 2
""
0 0
6 !z
" dxdzdy
0
r2 cos! dzdrd!
! 3 sin 2 " cos# d!d#d" +
%
1
2%
2
sin "
%
0
0
6
$$ $
! 3 sin2 " cos# d! d#d"