# Double integral transforming into polar coordinates

1. Jan 7, 2010

### 8614smith

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

By transforming to polar coordinates, evaluate the following:
$$\int^{a}_{-a}\int^{\sqrt{}{{a^2}-{x^2}}}_{-\sqrt{{a^2}-{x^2}}}dydx$$

2. Relevant equations

3. The attempt at a solution

I can get the right answer to this but only after guessing that the inner limits are between 0 and a, and the outer limits are between 0 and $$2\pi$$.

Can anyone tell me why these are the limits and how to get to polar limits from cartesian?

What i mean is, what is the 'a' all about? i can't find anything about it on the net, i only managed to do this question from a guess as it looked very similar to an example question in my notes but without the 'a'.

$$\int^{a}_{-a}\int^{\sqrt{{a^2}-{x^2}}}_{-\sqrt{{a^2}-{x^2}}}dydx$$ => $$\int^{2\pi}_{0}\int^{a}_{0}rdrd\theta=\int^{2\pi}_{0}\left[\frac{r^2}{2}\right]^{a}_{0}d\theta$$

$$=\int^{2\pi}_{0}\frac{a^2}{2}d\theta=\left[frac{{a^2}\theta}{2}\right]^{2\pi}_{0}={\pi{a^2}}$$

2. Jan 7, 2010

### rock.freak667

The 'y' limits are $\sqrt{a^2-x^2}$

thus $y=\sqrt{a^2-x^2}$ or $y^2=a^2-x^2 \Rightarrow x^2+y^2=a^2$

which is a circle centered at the origin with radius 'a'.

3. Jan 7, 2010

### 8614smith

but thats only 1 limit isn't it? if its the integral of the complete circle why is there a negative $$\sqrt{a^2}-{x^2}$$ term?

I would have thought it would only need that one limit,as it is the equation of the entire circle.

Or is that positive limit the semi-circle in the 1st and 2nd quadrant and the negative limit the semi-circle in the 3rd and 4th quadrant(if it is this could you explain it a bit better than i have? as i don't quite understand fully what i've written)

4. Jan 7, 2010

### rock.freak667

x2+y2=a2 is the equation of the entire circle.

y=√(a2-x2) represents the upper half of the circle.

y=-√(a2-x2) represents the lower half of the circle.

So to fully integrate the integral, you'd need to integrate 'y' between √(a2-x2) and -√(a2-x2), and integrate 'x' between a and -a.

5. Jan 7, 2010

### 8614smith

Ok i'm sort of getting it, but why is it that $${x^2}+{y^2}={a^2}$$ gives the graph of a circle and $$y=\sqrt{1-{x^2}}$$ gives the graph of only half the circle?

If you subtract $${x^2}$$ from both sides from the 1st equation and then square root both sides you get the 2nd equation so surely they would be the same graph?

6. Jan 7, 2010

### rock.freak667

for y=√(a2-x2), for every value of x<a, the value of y is positive. So this will always give values of y>0. So if you plot these values you will see that it forms a semi-circle.

7. Jan 7, 2010

### 8614smith

ah i see now, its the ± thing you get when you square root that makes the difference,
thanks!

8. Jan 7, 2010

### 8614smith

double integral polar coordinates...

1. The problem statement, all variables and given/known data
By transforming to polar coordinates, evaluate the following:
$$\int^{2}_{0}\int^{\sqrt{4-{x^2}}}_{\sqrt{y(2-y)}}\frac{y}{{x^2}+{y^2}}dxdy$$

2. Relevant equations

3. The attempt at a solution
Ive drawn the graph but can't work out how to get the limits as the r limits are not a radius around the origin, its between a sideways semi-circle centrered around (0,1) and $$x=\sqrt{2}$$ - a straight line, I could do it if i moved the semi circle to be centred around the origin and then subtract that area from the rectangle 2 x $$\sqrt{2}$$ but i'm not sure this is the way they want me to do it??

i've got this far but can't work out limits:

$$\int\int\frac{rsin\theta}{r^2}rdrd\theta=\int\int\frac{r^2}{r^2}\frac{sin\theta}{r^2}drd\theta=\int\int\frac{sin\theta}{r^2}drd\theta$$

9. Jan 7, 2010

### LCKurtz

Re: double integral polar coordinates...

I assume you have a typo in the upper limit on the inner integral. Shouldn't the inner integral look like:

$$\int^{2}_{0}\int^{\sqrt{4-{y^2}}}_{\sqrt{y(2-y)}}\frac{y}{{x^2}+{y^2}}dxdy$$

My next question is whether you were given that integral or whether those limits are your attempt at describing the region. Are you trying to describe the region in the first quadrant exterior to the sideways circle but inside the larger circle? I suspect your integral has more wrong than just that typo. Please supply the exact statement of the problem as stated in the text.

10. Jan 8, 2010

### 8614smith

Re: double integral polar coordinates...

hi, no thats exactly as it is in the question, is that not possible to do it like that then?

the answer given is $$2-{\pi/2}$$

Ive done the question assuming it was a typo and got the answer as $${\pi/2}$$ but ive split the integral into two double integrals the 1st being a circle of radius 2 and integrated in the 1st quadrant only, then subtracted the 2nd double integral of a semi-circle of radius 1 but ive moved the centre point to the origin as i don't know how to handle it as one double integral.

11. Jan 8, 2010

### HallsofIvy

Re: double integral polar coordinates...

No, that is not possible: if you integrate with respect to y and then put in a lower limit with y in it, you will still have a "y" in the function to be integrated with respect to x. Your result would be function of y, not a number. The lower limit must be $\sqrt{x(2- x)}$.

$y= \sqrt{4- x^2}$ is, as said before, the upper half of the circle with center at the origin and radius 2.

$y= \sqrt{x(2- x)}$ is the upper half of $y^2= x(2- x)= 2x- x^2$. That is the same as $x^2- 2x+ y^2= 0$ and completing the square give $(x- 1)^2+ y^2= 1$, the circle with center at (1, 0) and radius 1.

What happens now is that the smaller circle is completely contained in the larger. Frankly, the say I would do this is say that the area of the larger circle, with radius 2, is $\pi(2^2)= 4\pi$ and the area of the smaller circle, of radius 1, is $\pi(1^2)= \pi$ so the area "between" them is $4\pi- \pi= 3\pi$. Since we are only interested in the upper half of both of these, the area sought is $(3/2)\pi$.

12. Jan 8, 2010

### LCKurtz

Re: double integral polar coordinates...

With the change in the upper limit I gave you:

$$\int^{2}_{0}\int^{\sqrt{4-{y^2}}}_{\sqrt{y(2-y)}}\frac{y}{{x^2}+{y^2}}dxdy$$

you will get that answer. You have to remember that your limits in polar coordinates go from r on the inner curve to r on the outer curve, and in this case, theta goes from zero to pi/2. Also, contrary to the statements in some other posts, you are not computing an area since the integrand is not 1.

Your little half circle given by your lower limit is x = sqrt(y(2-y)) can be rewritten as

$$x^2 + (y-1)^2 = 1$$

You need to translate this to polar coordinates. You should be able to show that its polar equation becomes:

$$r = 2\sin\theta$$.

$$\int_0^{\frac{\pi} 2} \int_{2\sin\theta}^2 ...$$