Triple integrating cylindrical coordinates?

[blackdarkeye]

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!

 

[LCKurtz]

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.

 

[batmankiller]

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

 

[LCKurtz]

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.