Change the order of integration and polar coordinates?

Question

http://img3.freeimagehosting.net/image.php?6a6c1a8a03.jpg

Please help me with this problem. Thanks!!

I think i have the right picture but then when i switch from dydx to dxdy, I have trouble finding the parameters.
Also when you switch it to polar, will the equestion just be "r"?

Answer 1

∫(0 to 2) ∫(0 to √(2x-x^2)) √(x^2 + y^2) dydx
Note that 2x-x^2 is a parabola with zeroes at 0 and 2 (which are the limits of x) and maximum of 1 at x = 1. The y-limit is the square root of this, so we have y going from 0 to 1.

Now y = √(2x-x^2) => 2x - x^2 = y^2 => x^2 - 2x - 1 = 1-y^2 => (x-1)^2 = 1-y^2 => x = 1 ± √(1-y^2). So for the dxdy integral we get
∫(0 to 1) ∫(1-√(1-y^2) to 1+√(1-y^2)) √(x^2 + y^2) dxdy.

For the polar coordinates it becomes a bit tricky. To answer your question, no, the integrand will not merely be r - remember that in the change of coordinates we replace dx dy with r dr dθ, so the integrand will be r^2.
Note that x^2 + y^2 ≤ x^2 + (2x-x^2) = 2x, so r has a maximum of 2, and further as we travel from left to right along the curve r is monotonic increasing. This means that for a fixed value of r, we go from the positive x-axis to the curve in angle.

So for each value of r between 0 and 2 we go out to the value of θ determined by x = r^2/2, y = √(2x - x^2) = √(r^2 - r^4/4) = r√(1 - r^2/4). So we get
∫(0 to 2) ∫(0 to arctan((2/r)√(1-r^2/4))) r^2 dθdr.

Answer 2

you in straightforward terms could confirm what area you have been integrating over in cartesian coordinates and exhibit it in polar coordinates. x is going from 0 to a million, and y is going from 0 to sqrt(a million-x^2). Draw it out. pay specific interest to the better decrease of y: y = sqrt(a million-x^2) is the comparable subject as x^2+y^2=a million, it is the equation of a circle of radius a million based on the beginning. The area of integration, then, is the 1st quadrant of a circle. r will bypass from 0 to a million, and theta will bypass from 0 to pi/2.