Polar Coordinate Integration

1. Nov 8, 2013

PsychonautQQ

1. The problem statement, all variables and given/known data
Integrate the double Integral: 6xdydx in polar coordinates

The y goes from bottom limit of x(3)^(1/2) to the top limit of (1-x^2)^(1/2)
the x goes from 0 to 1/2

2. Relevant equations

3. The attempt at a solution
So I graphed it, and it looks like a semi circle on the positive y plane with a linear line going through it in the first domain.
Changing this to polar coordinates I got
Double Integral: 6r^2cosθdrdθ

and for dr I evaluated it between 0 and 1
for dtheta, i'm having trouble figuring it out. It looked to me like it should go from pi/4 to pi/2, because the radius is one and x goes from 0 to 1/2 only. Any advice?

2. Nov 8, 2013

physics&math

Are you sure you have the lower limit on y written correctly? Did you mean y = x3/2 or y = 3x1/2?

Either way, neither of those lines are linear. Which means that r isn't going from 0 to 1 but from the lower y limit (converted to polar) to 1.

To find the upper limit on theta, you have to find where the two equations intersect and the corresponding angle.

3. Nov 8, 2013

LCKurtz

I'm assuming the lower curve is $y = \sqrt 3 x$. What angle does that line make with the $x$ axis? That should give you a hint about the lower $\theta$ value. It would be good of you to state the exact problem or provide a figure. Does it ask for the region above the line and below the circle? Or above the line and above the $x$ axis and below the circle?? Or something else? Where does the $x=\frac 1 2$ come from?

Last edited: Nov 8, 2013