Triple integrating cylindrical coordinates?

blackdarkeye
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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

Homework Equations



x=rcos(\theta)
y=rsin(\theta)
r=sqrt(x^2+y^2)

The Attempt at a Solution



i converted it into cylindracl coordinates and got

\int\int\int -4rcos(\theta)+3rsin(\theta) r dzdrd\theta

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

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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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.
 
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
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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