Download Mat 241 Chapter 15 Test

Mat 241 Chapter 15 Test
2010 A & B KEY
Name___________________________________
Professor Schultz
#1. (5 pts) The shown integral is somewhat “evil” in the order of integration
specified. To overcome this obstacle, compute the exact integral by
reversing the order of integration.
 3 x2
ye
ye
ye
0 1 x8 dxdy  0 0 x8 dydx  0  3x8

y3
du
let u  x 2 ,  xdx
2
8 2
2 x3
2 x2
2 x2
2
x3
0
2
3

2
2
x2
x 3  ex

1xe
dx 
0 3x8 dx  0 3 dx


4
4
eu
eu
1
  dx 
  e4  1 
6
6 0 6
0
#1. (5 pts) The “sin-log” function has no known antiderivative. To overcome
this obstacle, compute the exact integral by reversing the order of
integration. Note: log(x) denotes the natural logarithm ln(x) for most
mathematicians, engineers, and computer languages. The notation ln is for
greenhorns.
e
sin  x 
0 y log  x dxdy  1
e
1 e
log x 

0
log( x )
e

sin  x 
sin  x  y
dx
dydx   

log  x 
log
x
  0 
1

e
 sin  x  log( x) sin  x   0 
e
 

dx   sin  x dx   cos  x  1
log  x 
log  x  
1
1
  cos  e   cos 1  cos 1  cos  e 
e
#2. (10 pts) Using a double or triple integral to find the volume of the solid
tetrahedron with vertices:
 0,0,0  , 0,0,1 ,  0,2,0  , and 2,2,0  .
To determine the plane’s equation we do the following:





1. Select a vertex from: 0,0,1 , 0,2,0 ,or 2,2,0 .
2. Create two vectors using the selected point and the other two
remaining points.
3. Utilizing those two vectors we compute the cross-product.
4. With the normal vector and one of the three points we determine the
plane’s equation upon which we can integrate to get our desired
volume.
P  0,0,1  ,Q  0,2,0  ,and R 2,2,0 
PQ  2,0,0 ;PR  2, 2,1
i
PQ  PR  2
j
k
0
0   0  0  i   2  0  j   4  0  k  0,2,4
2 2 1
The plane's equation is:
0  x - 0   2  y  0   4  z  1   0  2y  4z  4  0
z 1
2 2
V  
y
2
1
y
2

0 x 0
2
2
2
 

y
y2 
x2  

1dydx    1  dydx    y   dx    1   x   dx
2
4 x
4 

0 x
0
0
2 2
2



x2
x3 x2 
8 4 8 2
   1   x dx   x     2    
4
12 2  0
12 2 12 3


0
2
Of course, we could just use the 7th grade formula:
Vpyramid 
1
1 1
2
 area_ of _ the _base height    2  2   1 
3
3 2
3

#3. (5 pts) Evaluate the following integral exactly.
A perfect candidate for polar coordinates.

 x2
 
 
sin  x 2  y 2 dydx
0
 
   sin r2 rdrd let u  r2 ;
0 0
 

du
 rdr
2


1
1

  sin udud    cosu 0 d     1  1 d   d  
20
20
0 0
0
#4. (6 pts) For the pictured solid assume that the density is proportional to
the distance to the xz - plane. The solid is formed by the following
surfaces:
y  x, z  1  x2 , x  1, y  2x
A. Set – up ONLY! An integral to compute the mass of the solid in the
dydzdx order. ALSO sketch the plane projection.
1 1 x2 2x
mass  
  kydydzdx
0
2
0
x
z
1.8
1.6
1.4
1.2
1
0.8
z  1x2
0.6
0.4
0.2
x
-0.8
-0.6
-0.4
-0.2
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
-0.2
B. Set – up ONLY! An integral to find
sketch the plane projection.
M xz
in the dzdydx order. ALSO
1 2x 1 x2
Mxz   
0 x
y

ky 2dzdydx
0
4.5
4
3.5
3
2.5
2
1.5
R
1
yx
0.5
x
-0.8
-0.6
-0.4
-0.2
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
I y in the dydxdz order. ALSO
C. Set – up ONLY! An integral to find
sketch the plane projection.
1 1z2 2x
Iy  
0
  x
0
2
 z2 kydydxdz
x
2
z
1.8
1.6
1.4
1.2
1
0.8
z  1x2
0.6
0.4
0.2
x
-0.8
-0.6
-0.4
-0.2
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
-0.2
#5. (10 pts) FIND THE VOLUME of the solid that is bounded between the
two cones given by:
z  x 2  y2 and
z  3 x2  y2 
And between the two hemi-spheres:
z  4  x 2  y2 and
z  1  x2  y2
To help you, the projection in the yz-plane is shown below. The area to be
swept into a volume is shown as
z
z  3x2  y2 
z  4 x2  y2
z  x2  y2
z  1  x2  y2
y
Looks like a case for spherical coordinates to me! I’ll need to find ranges
for , ,and  .
From the picture:
zinnersphere  1  x 2  y2  x 2  y 2  zoutsidecone
1  r2  r2  r 
1
2
and
zoutersphere  4  x 2  y 2  3  x 2  y 2   zinsidecone
4  x2  y2  3  x2  y2 
4  r2  3r2  r  1
Our first phi angle is:
z
z  3x2  y2 
z  4 x2  y2
z  x2  y2
1
2
z  1x y
2
2


6
y
Our second phi angle is:
z
z  3x2  y2 
z  4 x2  y2
z  1  x2  y2
z  x2  y2
1/sqrt(2)


4
1
y
So we have:


1    2,    ,0    2
6
4
V

2 4 2

2 4
0  1
6
6
2
   sin ddd 

0   3


2 4

7
 sin dd    sin dd

30
1
3 2
6
2
2
2

7
7 
7  2
3

 
4
   cos   d     cos    cos     d    

 d
30
3
4
6
3
2
2






6
0
0 

7

6

3 2
2
  d  73 
3 2

0
#5. (10 pts) Calculate the exact volume of the child’s plastic chair formed
by the intersection of
x2  y2  1 ,
2
2
the two surfaces z  x  5 and z  4  y
the cylinder
and having
y0
(Note: The chair has been rotated so you can see inside it.)
Let x and y equal zero to determine top and bottom surfaces.
z  x 2  5 and z  4  y 2
when x  y  0
z  5 & z  4
We are integrating over the circle x 2  y 2  1 with y  0.
1
V 
1
1  x2 4  y2
 
0
x 5
2
 1 4 r2 sin2 
1dzdydx  

rdzdrd
0 0 r cos  5
2
2
 1
   4  r2 sin2    r2 cos2   5  rdrd
0 0
 1
  9  r2 sin2   r2 cos2  rdrd
0 0
 1
  9  r2 rdrd
0 0
 1
  9r  r 3 drd
0 0
1

 9r2 r 4 
17
17
  d    d 
 
4
4 0
4 0
2
0
