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1
Convert point P(1, 3, 5) from
Cartesian to cylindrical and
spherical coordinates.
Solution
2
3
2 2
2 2
1, 3, 5
(1) 3 10
3.16
Given
x y z
x y

  
  
  

4
1 1
1 0
0
1, 3, 5
3
tan tan
1
tan (3) 72
( , , ) (3.16,72 ,5)
Given
x y z
y
x
z

 
 

  
  
 
 
2 2 2
1, 3, 5
1 9 25 35
5.92
Given
x y z
r x y z
  
   
   

5
2 2
1
1 1
0
0 0
1, 3, 5
tan
10
tan tan 0.6324
5
32.5
( , , ) (5.92,32.5 ,72 )
Given
x y z
x y
z
r

 

 
  

 
 

 
Convert point T(0, -4, 3) from
Cartesian to cylindrical and
spherical coordinates.
Solution
2 2
2 2
0, 4, 3
(0) ( 4) 16
4
Given
x y z
x y

   
  
   

6
1 1
1 0
0
0, 4, 3
4
tan tan
0
tan ( ) 270
( , , ) (4,270 ,3)
Given
x y z
y
x
z

 
 

   

  
  
 
2 2 2
0, 4, 3
0 16 9 25
5
Given
x y z
r x y z
   
   
   

7
2 2
1
1 1
1 0
0 0
0, 4, 3
tan
16 4
tan tan
3 3
tan 1.33 53.5
( , , ) (5,53.5 ,270 )
Given
x y z
x y
z
r

 

 

   

 
 
 
 
Convert point S(-3, -4, -10) from
Cartesian to cylindrical and
spherical coordinates.
Solution
8
2 2
2 2
3, 4, 10
( 3) ( 4) 25
5
Given
x y z
x y

     
  
    

1 1
1 0
0
3, 4, 10
4
tan tan
3
tan (1.33) 233.5
( , , ) (5,233.5 , 10)
Given
x y z
y
x
z

 
 

     

  

 
  
9
2 2 2
3, 4, 10
9 16 100 125
5 5 11.18
Given
x y z
r x y z
     
   
   
 
2 2
1
1 1
0
0 0
3, 4, 10
tan
25
tan tan ( 0.5)
10
153.5
( , , ) (11.18,153.5 ,233.5 )
Given
x y z
x y
z
r

 

 
     

 
  


 

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Cartesian.doc

  • 1. 1 Convert point P(1, 3, 5) from Cartesian to cylindrical and spherical coordinates. Solution
  • 2. 2
  • 3. 3 2 2 2 2 1, 3, 5 (1) 3 10 3.16 Given x y z x y           
  • 4. 4 1 1 1 0 0 1, 3, 5 3 tan tan 1 tan (3) 72 ( , , ) (3.16,72 ,5) Given x y z y x z                 2 2 2 1, 3, 5 1 9 25 35 5.92 Given x y z r x y z            
  • 5. 5 2 2 1 1 1 0 0 0 1, 3, 5 tan 10 tan tan 0.6324 5 32.5 ( , , ) (5.92,32.5 ,72 ) Given x y z x y z r                  Convert point T(0, -4, 3) from Cartesian to cylindrical and spherical coordinates. Solution 2 2 2 2 0, 4, 3 (0) ( 4) 16 4 Given x y z x y             
  • 6. 6 1 1 1 0 0 0, 4, 3 4 tan tan 0 tan ( ) 270 ( , , ) (4,270 ,3) Given x y z y x z                    2 2 2 0, 4, 3 0 16 9 25 5 Given x y z r x y z             
  • 7. 7 2 2 1 1 1 1 0 0 0 0, 4, 3 tan 16 4 tan tan 3 3 tan 1.33 53.5 ( , , ) (5,53.5 ,270 ) Given x y z x y z r                     Convert point S(-3, -4, -10) from Cartesian to cylindrical and spherical coordinates. Solution
  • 8. 8 2 2 2 2 3, 4, 10 ( 3) ( 4) 25 5 Given x y z x y                 1 1 1 0 0 3, 4, 10 4 tan tan 3 tan (1.33) 233.5 ( , , ) (5,233.5 , 10) Given x y z y x z                      
  • 9. 9 2 2 2 3, 4, 10 9 16 100 125 5 5 11.18 Given x y z r x y z                 2 2 1 1 1 0 0 0 3, 4, 10 tan 25 tan tan ( 0.5) 10 153.5 ( , , ) (11.18,153.5 ,233.5 ) Given x y z x y z r                      