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(MATH100)quiz02sol-1b.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: T1B
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
Z 2 Z 1 Z 4.x.y
1 dz dx dy.
000
Solution: The solid G described by 0 . y . 2, 0 . x . 1, 0 . z . 4 . x . y is sketched in the following. Remark that since x, y, z are all positive, the .gure must lie in the .rst octant. The top surface of the solid is nothing but just the plane z =4 . x . y, or x + y + z = 4. z
4
x + y + z =4
The solid G
y
2
1
x
2. (3 Marks) Evaluate the double iterated integral by interchanging the order of integration.
Z 4 Z 2
cos(2y 3) dy dx.
0 x
Solution: We cannot do this integral by integrating with respect to y .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 I :0 . x . 4 and x . y . 2.
Here is a sketch of this region (left).
yy
y =2 2 x =0 x = y 2 y = x
xx
0 40
So, if we reverse the order of integration we get the following limits (sketch above on the right). Type II :0 . y . 2 and 0 . x . y 2 . The integral becomes
2
Z 4 Z 23Z 2 Z y 3
cos(2y ) dy dx = cos(2y ) dx dy
0 x 00
2
3iy
= Z 2 hx cos(2y )dy
0 Z 2 0
= y 2 cos(2y 3) dy 0
1 Z 2 33
= cos(2y ) d(2y )
6
0 1 3i2
= hsin(2y ),
6 0 or
Z 4 Z 23
cos(2y ) dy dx = sin 16.
0 x
3. (2 Marks) Use polar coordinates to evaluate the iterated integral
1.y
Z 1 Z 2
p1 . x2 . y2 dx dy. 00
Solution: By the limits of the given integral we know that the domain of integration is
2
0 . y . 1, 0 . x . p1 . y.
The above region is nothing but one quarter of a unit circle centered at the origin in the .rst quadrant. In polar coordinates the region can be represented by
0 . r . 1, 0 . . .
2 The iterated integral is then
Z 1 Z 1.y2 p1 . x2 . y2 dx dy = Z /2 Z 1 p1 . r2 r dr d
0 0 0 0
= (. 1 ) .(1 . r 2)3/2 C1 = ( )(. 1 )( 2 )(0 . 1),
2 2 3/2 0 2 2 3
or
Z 1 Z 1.y2 p1 . x2 . y2 dx dy = .
0 0 6
222
4. (3 Marks) Let G be the solid region inside the sphere x + y + z = 6 and above the paraboloid z = x 2 +y 2 . Express the volume of the solid 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 = x 2 + y 2
The solid G
222
x + y + z =6
2
y
x
We use cylindrical coordinates for this problem. Recall that
x = r cos , y = r sin , z = z
are the conversion formulas from rectangular coordinate