How Do I Convert This Integral's Domain to Polar Coordinates?

[frggr]

Alright, so I'm having a problem converting to polar co-ordinates
Never been confused by polar before :(I'm trying to convert the domain of an integral to polar co-ordinates

The domain would be something like

D:{(x,y)| 1<x<2 , 0<y<(2x-x²) ^1/2}

First I recognize that y=(2x-x²) ^1/2 is the equation for a circle of radius shifted 1 unit to the right

so

y² + (x-1)² = 1

and it is the domain from x = 1 to x =2 (So the top right quadrant of the circle)

[PLAIN]http://img202.imageshack.us/img202/4919/graphua.png

Graphically I can deduce that since x = y = 1 at the peak, then cos$$\vartheta$$ = sqrt(2) :. $$\vartheta$$ = $$\frac{\pi}{4}$$

and it looks like r goes from 0 to 2cos$$\vartheta$$ //Which I am told is wrong

This leads me to two questions:
1) How would I find r graphically here?
2) Is it possible to convert to polar algebraically?

I tried to convert the inequalities seperately but that led me to
D:{(r,$$\vartheta$$)| $$\frac{\pi}{4}$$ < $$\vartheta$$ < 0, 0 < r < 2cos$$\vartheta$$}

//My method is below
0<y<(2x-x²)<sup>1/2
02 < y2 < 2x - x2
0 < y2 + x2 < 2x
0 < r2 < 2rcos[tex]\vartheta[\tex]
0/r < r < 2cos$$\vartheta$$

1 < x < 2
1 < rcos$$\vartheta$$ < 2
r has a maximum value of 2cos$$\vartheta$$
1 < 2 cos2$$\vartheta$$ < 2
1/2 < cos2$$\vartheta$$ < 1
1/sqrt(2) < cos$$\vartheta$$ < 1
$$\frac{\pi}{4}$$ < $$\vartheta$$ < 0</sup>

 

[Mark44]

Alright, so I'm having a problem converting to polar co-ordinates
Never been confused by polar before :(


I'm trying to convert the domain of an integral to polar co-ordinates

The domain would be something like

D:{(x,y)| 1<x<2 , 0<y<(2x-x²) ^1/2}

First I recognize that y=(2x-x²) ^1/2 is the equation for a circle of radius shifted 1 unit to the right

so

y² + (x-1)² = 1

and it is the domain from x = 1 to x =2 (So the top right quadrant of the circle)

[PLAIN]http://img202.imageshack.us/img202/4919/graphua.png

Graphically I can deduce that since x = y = 1 at the peak, then cos$$\vartheta$$ = sqrt(2) :. $$\vartheta$$ = $$\frac{\pi}{4}$$

and it looks like r goes from 0 to 2cos$$\vartheta$$ //Which I am told is wrong

This leads me to two questions:
1) How would I find r graphically here?
2) Is it possible to convert to polar algebraically?

I tried to convert the inequalities seperately but that led me to
D:{(r,$$\vartheta$$)| $$\frac{\pi}{4}$$ < $$\vartheta$$ < 0, 0 < r < 2cos$$\vartheta$$}

//My method is below
0<y<(2x-x²)<sup>1/2
02 < y2 < 2x - x2
0 < y2 + x2 < 2x
0 < r2 < 2rcos[tex]\vartheta[\tex]
0/r < r < 2cos$$\vartheta$$

1 < x < 2
1 < rcos$$\vartheta$$ < 2
r has a maximum value of 2cos$$\vartheta$$
1 < 2 cos2$$\vartheta$$ < 2
1/2 < cos2$$\vartheta$$ < 1
1/sqrt(2) < cos$$\vartheta$$ < 1
$$\frac{\pi}{4}$$ < $$\vartheta$$ < 0</sup>

^{Your last inequality is wrong - theta is between 0 and pi/4, but your inequalilty says that pi/4 < 0.
The range for theta should be
$$0 \leq \theta \leq \frac{\pi}{4}$$}

 

[frggr]

I know, that's one of the problems. The inequalities didn't switch when I tried to do it algebraicaly :. I'm doing something wrong. The first inequality wrong as well. it should be
$$\frac{1}{cos \theta} \leq r \leq 2cos \theta$$

 

[Mark44]

Yes, you are doing something wrong. Look at the graph of y = cos(theta) on the interval [0, pi/2]. The function is decreasing on this interval, so if the y values are 1/sqrt(2) < cos(theta) < 1, the theta values will be in the opposite order. I.e., 0 < theta < pi/4.