# Problem with limits of integration - converting double integral to polar form

1. Dec 5, 2012

### e^(i Pi)+1=0

1. The problem statement, all variables and given/known data
$\int_0^2 \int_0^\sqrt{2x-x^2} xy,dy,dx$

I know the answer, but how does the 2 in the outer integral become pi/2?? I'm fine with everything else, I just can't get this...

2. Dec 5, 2012

### clamtrox

Do you have a particular reason for calculating this integral in polar form? It's much easier with Carteesian coordinates. Also I don't understand your question. What is "outer integral"?

3. Dec 5, 2012

### e^(i Pi)+1=0

The outer integral is the dx integral, and it's a practice problem out of the book. I know the upper limit for dy in polar is r=2cos(theta) which I got by taking y=root(2x-x2) and converting to polar and solving for r. Why won't this work for dx?

4. Dec 5, 2012

### clamtrox

But what good is that? You need the limits for r and theta, not for x and y. If you want to change to polar coordinates, start by figuring out how x and y are related to r and θ (hint, x≠rcosθ)

5. Dec 5, 2012

### e^(i Pi)+1=0

I don't think you read very carefully.

6. Dec 5, 2012

### haruspex

You mean, the upper limit for r is 2cos(θ).
To understand the range for θ, it's easiest to draw a picture. The region is bounded above by an inverted parabola passing through (0,0) and (2,0). Since it is wholly within the first quadrant, you have 0 ≤ θ ≤ π/2. For any given θ in that range, r can be anything from 0 to the point on the parabola at that θ. Allowing θ to go from 0 to π/2 therefore captures the whole region and nothing but.

7. Dec 6, 2012

### HallsofIvy

Staff Emeritus
Well said, haruspex, but the area is not an "inverted parabola". It is the upper semi-circle of the circle with center at (1, 0) and radius 1.

$y= \sqrt{2x- x^2}$ and, squaring, $y^2= 2x- x^2$.
$x^2- 2x+ y^2= 0$, $x^2- 2x+ 1+ y^2= 1$,
$(x- 1)^2+ y^2= 1$.

8. Dec 6, 2012

### haruspex

Good catch. I must have dropped the square root.