# Homework Help: Double integral: Cartesian to Polar coordinates

1. Dec 13, 2013

### IsaacStats

1. The problem statement, all variables and given/known data

∫∫√(x^2+y^2)dxdy with 0<=y<=1 and -SQRT(y-y^2)<=x<=0

2. Relevant equations

x=rcos(theta)
y=rsin(theta)

3. The attempt at a solution

0.5<=r=1, we get r=0.5 from -SQRT(y-y^2)<=x by completing the square on the LHS
then, 0<=theta<=pi

But, when I calculated the original integral, I got 0.2222...
When I calculate the integral with the polar coordinates, I get 0.92

2. Dec 13, 2013

### ShayanJ

Maybe you forgot the determinant of the Jacobian?
Anyway,without taking a look at your calculations,we can't help!

3. Dec 13, 2013

### IsaacStats

Last edited: Dec 13, 2013
4. Dec 13, 2013

### Bazinga55

U need a factor of r. da=rdrdtheta

5. Dec 13, 2013

### CAF123

Did you try to plot the region of integration? How did you get those limits?

6. Dec 13, 2013

### IsaacStats

Explanation:

I started out with the graph. In the first quadrant we have a quarter of a circle x^2+y^2=1 (radius = 1. In the second quadrant we have x^2+(y-0.5)^2=0.25 (radius=0.5). Then, since the function exists both in first and second quadrant, 0<=theta<=pi. Then, by my false reasoning, I made a mistake of thinking that 0.5<=r<=1.

After revising my calculations, I got 0<=theta<=pi/2, 0<=r<=1. I derived the limits for theta in the following way
0<=rsin(theta)<=1
-SQRT(rsin(theta)-(rsin(theta))^2)<=rcos(theta)<=0
rsin(theta)-(rsin(theta))^2=(rcos(theta))^2
sin(theta)=r
Then, by substituting into the first line, we have
0<=(sin(theta))^2<=1
Then, theta1=0, theta2=pi/2
Hence, r1=0, r2=1

7. Dec 13, 2013

### CAF123

Those bounds describe a portion of the unit circle centered on the origin in the first quadrant.

8. Dec 13, 2013

### IsaacStats

So, to get to the right conclusion do I add the area of the half of the circle centered at (0,0.5)?

9. Dec 13, 2013

### CAF123

Are you sure you are considering the right half of the curve x = ±√(y-y2)? For sure, this exists in both first and second quadrants, but you are only interested in the half x = -√(y-y2). Since √(y-y2) ≥ 0, what quadrant does x=-√(y-y2) lie?

10. Dec 13, 2013

### IsaacStats

In the second quadrant = > (pi/2)<=theta<=pi
0<=r<=0.5

11. Dec 13, 2013

### CAF123

I agree on the bounds for θ, but the values for r suggest a circle of radius 1/2 centered on the origin. Consider the curve x2+y2=y. Convert this to polar coordinates and you will obtain r. This value of r will depend on θ.

12. Dec 13, 2013

### IsaacStats

Thank you for correction. If (pi/2)<=theta<=pi, does it meant that sin(pi)<=r<=sin(pi/2); 0<=r<=1.

that's weird

13. Dec 13, 2013

### CAF123

These bounds say that as θ goes from π/2 to π, r goes from 0 to 1. r is measured from the origin and so r is precisely 1 at only one value of θ, namely θ = π/2. (See this from a sketch). So the bounds for r have to depend on θ, otherwise you are describing a portion of the unit circle centered at the origin.

14. Dec 13, 2013

### IsaacStats

That's great!

Does this mean that we can write the integral as a sum of two integrals (first quadrant circle and second quadrant circle)
0<=theta<=n/2, 0<=r<=1 + n/2<=theta<n, 0<=r<=sin(theta)
?

OR( 0<theta<n (0<=r<=1 + 0<=r<=sin(theta))

Last edited: Dec 13, 2013
15. Dec 13, 2013

### CAF123

Why did you tag on the part '0<=theta<=n/2, 0<=r<=1'? Remember x ≤ 0 so there should be nothing that describes anything in the first quadrant.

The bounds 'n/2<=theta<n, 0<=r<=sin(theta)' describe the required region. Did you draw a sketch? Convince yourself that this is the correct range for r - try out a few values of θ and see that you get reasonable values for r (r should monotonically decrease to 0 as θ sweeps from π/2 to π)

16. Dec 13, 2013

### IsaacStats

RIGHT! Function is undefined for X>0.

Therefore, n/2<=theta<=n, and 0<=r<=sin(theta)

17. Dec 13, 2013

### CAF123

Yes, everything makes sense?

18. Dec 13, 2013

### IsaacStats

Out of curiosity, why does not r go between 0 and csc(theta)?

19. Dec 13, 2013

### CAF123

The curve in Cartesians was $x^2 + y^2 = y$. Converting this to polar coordinates gives $r^2 = r\sin\theta$ from which we see that $r = \sin \theta$. $\theta = \pi/2$ is the maximum value for $r$, while $\theta = \pi$ is the minimum. In between, $r$ takes values given by the sine of the angle $\theta$.

20. Dec 13, 2013

### IsaacStats

EXCELLENT! THANKS!

I believe my mistake was looking at 0<=rsin(theta)<=1.

Thank You for your help and may numbers be in your favor.