Integrating a Semi-Circular Region in Polar Coordinates

[PsychonautQQ]

Homework Statement

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

Homework Equations

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?

 

[physics&math]

Are you sure you have the lower limit on y written correctly? Did you mean y = x^3/2 or y = 3x^1/2?

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.

 

[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?