Triple integrating cylindrical coordinates?

  • #1

Homework Statement



Integrate the function f(x,y,z)=−4x+3y over the solid given by the figure below, if P = (5,1,0) and Q = (-5,1,2).

[PLAIN]http://img259.imageshack.us/img259/958/sfig1681g1.gif [Broken]


Homework Equations



x=rcos([tex]\theta[/tex])
y=rsin([tex]\theta[/tex])
r=sqrt(x^2+y^2)


The Attempt at a Solution



i converted it into cylindracl coordinates and got

[tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] -4rcos([tex]\theta[/tex])+3rsin([tex]\theta[/tex]) r dzdrd[tex]\theta[/tex]

limits were arctan(-1/5)<[tex]\theta[/tex]<arctan(1/5), 0<z<2, 0<r<[tex]\sqrt{26}[/tex]

after integrating i got

-(8/3)(26^(3/2))sin(atan(1/5))-(2)(26^(3/2))cos(atan(1/5))+(8/3)(26^(3/2))sin(atan(-1/5))+(2)(26^(3/2))cos(atan(-1/5))

but apparently that is wrong. I need help!
 
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Answers and Replies

  • #2
LCKurtz
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Your setup looks good. I didn't work out the integral to see if you are correct or not, but I wonder if you are aware that you can simplify your sine and cosine of the arctan angles. Draw a picture of those two angles and put in their sine and cosine in your answer. Maybe it is correct and you don't know it. Or maybe not.
 
  • #3
i have the same problem and I can't seem to get the right answer .. i'm entering the answer in so it doesn't really matter if it's simplified or not
 
  • #4
LCKurtz
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Upon looking more carefully, your theta limits are not correct. Rember that arctan(x) returns a negative angle if x < 0. Your angle needs to go from the first quadrant to the second. So your lower limit would be arctan(1/5) and your upper limit would be pi - arctan(1/5).

Maple is giving an answer of 1040 if I didn't make any mistakes.
 

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