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(MATH100)quiz02sol-1a.pdf
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MATH 100 (L1) Introduction to Multivariable Calculus Spring 2006-07
. Quiz 02 Solutions: Multiple Integrals Duration: 40 minutes
Name: ID No.: Tutorial Section: T1A
Answer ALL of the following 4 short answer questions. Show all your work for full credit.
1. (2 Marks) Sketch the solid G whose volume is given by the triple integral
1.x
Z 1 Z 2 Z 2
1 dy dz dx.
00 0
Solution: The solid G described by
0 . x . 1, 0 . z . p1 . x2 , 0 . y . 2
is sketched in the following. In the .gure the cross-section curve (a portion of the unit circle in the
.rst quadrant of the xz-plane) is obtained by squaring z =1 . x2 and taking z positive. z
The solid G
1
Cross-section
x 2 + z 2 =1
y
2
1
x
2. (3 Marks) Evaluate the double iterated integral by interchanging the order of integration.
Z 1 Z 1
px3 +1 dx dy.
0 y
Solution: We cannot do this integral by integrating with respect to x .rst so we will hope that by interchanging the order of integration we will get something that can be integrable. Here are the limits for the variables that we get from the given integral.
Type II :0 . y . 1 and y . x . 1.
Here is a sketch of this region (left).
yy
2
y = x
x = yx =1
xx
00 y =0 1
So, if we reverse the order of integration we get the following limits (sketch above on the right).
Type I :0 . x . 1 and 0 . y . x 2 .
The integral becomes
Z 1 Z 1 Z 1 Z x 2
px3 +1 dx dy = px3 +1 dy dx
0 y 00
Z 1 ix = hypx3 +12 dx
0
0
Z 1 = x 2 px3 +1 dx 0
1 Z 13 1/23
=(x + 1)d(x + 1)
3
0 1 .(x 3 + 1)3/2 C1 = ,
33/2 0
or
Z 1 Z 1 2
px3 +1 dx dy = 23/2 . 1 .
9
0 y
3. (2 Marks) Use polar coordinates to evaluate the double integral ZZ 1
dx dy,
(1 + x2 + y2)2
D
where D = {x 2 + y 2 . a 2 : x . 0,y . 0}.
Solution: In polar coordinates the region D can be represented by
0 . r . a, 0 . . .
2 The double integral is then
Z /2
ZZ 1 Z a 1
dx dy = r dr d
(1 + x2 + y2)2 00 (1 + r2)2
D
Ca
.11 . 11 .
= . 2 = . 2 ,
2 21+ r4 1+0 1+ a
0
or
ZZ 1 a2
dx dy =.
(1 + x2 + y2)2 4(1 + a2)
D
4. (3 Marks) Let G be the solid in the .rst octant bounded by the cylinders x 2 + y 2 = 1 and x 2 + y 2 = 6 and the planes z = 0, z = 1, x = 0, and x = y. Express the volume of the solid G as an iterated triple integral in cylindrical coordinates. You are NOT required to evaluate the triple integral but you have to write the integration limits properly.
Solution: For reference purpose we make a sketch of the solid G in the following.
z
z =1
r =1
r =6
The solid G
1
6
4 y
y = x
x
We use cylindrical coordinates for this problem. Recall that
x = r cos , y = r sin , z = z
are the conversion formulas from rectangular coordinates to cylindrical coordinates. We can easily express the equations of cylinders in terms of cylindrical coordinates. They are
r =1,r =6.
In terms of cylindric