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(MATH100)quiz02sol-1c.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: T1C
Answer ALL of the following 4 short answer questions. Show all your work for full credit.
1. (2 Marks) Suppose that the area of a region in polar coordinate system is given by the double integral
Z 3/4 Z 2 sin
r dr d,
/4 csc
where csc =1/ sin . Sketch and shade the region and .nd its area.
Solution: By the limits of the given integral we know that the region (say R) is described by

3 1
. . , . r . 2 sin .
4 4 sin
First we should identify what are the corresponding graphs of the polar equations r =1/ sin and r = 2 sin . In fact, r =1/ sin represents a horizontal straight line (i.e., y = 1) whereas r = 2 sin represents a circle of radius 1 centered at (0, 1) in the xy-plane (i.e., x 2 +(y . 1)2 = 1). The region R is sketched in the following.
The region R
=3/4 = /4
1
r =
sin
r = 2 sin Polar axis
O
Obviously, the area of the region is given by
1
area = (1)2 =.
22
2. (3 Marks) Evaluate the double iterated integral by interchanging the order of integration.
Z 2 Z 1 x 3
ye dx dy.
0 y/2
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.
y
Type II :0 . y . 2 and . x . 1.
2
Here is a sketch of this region (left).
yy
y =2x x = y/2 x =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 . 2x.
The integral becomes
Z 2 Z 1 3 Z 1 Z 2x 3
ye x dx dy = ye x dy dx 0 y/2 00
Z 1 .1 2C2xx 3
= y edx
2
00
Z 1 3 .2 3 C1
2 xx
=2xe dx = e,
3 or
00
3 2
Z 2 Z 1 x
ye dxdy =(e . 1).
3
0 y/2
3. (2 Marks) The rectangle R in the xy-plane consists of those points (x, y) for which 0 . x . 2 and 0 . y . 1. Find the volume of the solid that lies below the surface z =1+ xy and above R.
Solution: The volume is given by the double integration.
ZZ
Volume = z dA
R
Z 2 Z 1
= (1+ xy) dy dx
00 Z 2 . 1 2C1
= y + xy dx
2 Z 2 .1 .. 1 2C2
00
= 1+ x dx = x + x,
24
or

00
volume of the solid = 3.
4. (3 Marks) Let G be the solid that is bounded below by the cone z = p(x2 + y2)/3 and bounded
222
above by the sphere x + y + z = z. Express the volume of the solid G as an iterated triple integral in spherical coordinates. You are NOT required to evaluate the triple integral but you have to write the integration limits properly.
Solution: We use spherical coordinates for this problem. Recall that
x = sin cos , y = sin sin , z = cos
are the conversion formulas from rectangular coordinates to spherical coordinat