# Polar Coor Intergration Help Part Duex

• Nov 10th 2009, 07:02 PM
TheUnfocusedOne
Polar Coor Intergration Help Part Duex
Hey again guys
for some reason I cant get this to work again.

Evaluate the integral by changing to polar coordinates.
• Nov 10th 2009, 08:20 PM
HallsofIvy
Quote:

Originally Posted by TheUnfocusedOne
Hey again guys
for some reason I cant get this to work again.

Evaluate the integral by changing to polar coordinates.

First step, draw a picture. x lies between 0 and 1/2 and, for each x, y is between the straight line $\displaystyle y= \sqrt{3}x$ and the unit circle $\displaystyle x^2+ y^2= 1$.

In polar coordinates, the unit circle is given, of course, by r= 1 and, since the slope of a line is the tangent of the angle the line makes with the x-axis, the line $\displaystyle y= \sqrt{3} x$ makes an angle [tex]\pi/3[tex]. Further, that line just happens to cross the circle where $\displaystyle x^2+ y^2= x^2+ (\sqrt{3} x)^2= 4x^2= 1$: x= 1/2.

The region in this integral s the segment of the unit circle in which r goest from 0 to 1 and $\displaystyle \theta$ goes from $\displaystyle \pi/3$ to $\displaystyle \pi/2$.
• Nov 10th 2009, 08:35 PM
TheUnfocusedOne
r goes from 0 to 1?
you gotta be kidding me....
thanks a lot