# Changing the order of integration

1. Feb 23, 2009

### a1010711

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

change the order of integration in the follwing double integral:

intgral [0,a] integral [0,sqrt(2ay-y^2)] f(x,y) dx dy

if x= sqrt(2ay-y^2),, do i solve for y or something,,ive used the computer to graph it but that didnt help me.

2. Feb 23, 2009

### HallsofIvy

Staff Emeritus
Then presumably you see, from the graph, that the region of integration is a semi-circle with center at (0,a), radius a and extending to the right of the y-axis.

If you had not graphed it you could still get that by noting that $x= \sqrt{2ay- y^2}$ can be expanded, by squaring both sides, to $x^2= 2ay- y^2$ or $x^2+ y^2- 2a= 0$. Completing the square, $x^2+ (y- a)^2= a^2$ which is a circle of radius a, center at (x, 0). Of course the square root is never negative so that is just the semi-circle with x non-negative.

However you determine region, to reverse the order of integration, so that you are integrating with respect to y first, and then with respect to x, the limits of integration on x must be numbers so they should be the smallest and largest possible values of x in that semi-circle. To get the limits of integration for y, draw a vertical linel, representing a particular value of x. The limits of integration for y are the lowest and highest value of y on that line- the values at the endpoints (and, of course, they depend on x).