# Thread: change of coordinates

1. ## change of coordinates

i tried to use Latex but i failed so here's the problem the double integral of (x^2 + y^2) ^1/2 dx dy the inner limit is from 0 to (3y-y^2)^1/2 and the outer limit is from 0 to 3

i know how to change the coordinates but having problem figuring out the limits for this problem

2. Originally Posted by sean123
i tried to use Latex but i failed so here's the problem the double integral of (x^2 + y^2) ^1/2 dx dy the inner limit is from 0 to (3y-y^2)^1/2 and the outer limit is from 0 to 3

i know how to change the coordinates but having problem figuring out the limits for this problem
$\int_{y= 0}^3\int_{x= 0}^{3y- y^2} \sqrt{x^2+ y^2} dxdy$
$x= 3y- y^2$ is a parabola with horizontal axis. It crosses the y-axis at (0, 0) and (0, 3). Completing the square, $x= \frac{9}{4}- (y- \frac{3}{2})^2$ so that $y- \frac{3}{2}= \pm\sqrt{x- \frac{9}{4}}$. That tells us that x ranges from 0 to $\frac{9}{4}$ and that, for each x, y ranges from $\frac{3}{2}- \sqrt{x- \frac{9}{4}}$ to $\frac{3}{2}+ \sqrt{x- \frac{9}{4}}$.

3. Originally Posted by HallsofIvy
$\int_{y= 0}^3\int_{x= 0}^{3y- y^2} \sqrt{x^2+ y^2} dxdy$
$x= 3y- y^2$ is a parabola with horizontal axis. It crosses the y-axis at (0, 0) and (0, 3). Completing the square, $x= \frac{9}{4}- (y- \frac{3}{2})^2$ so that $y- \frac{3}{2}= \pm\sqrt{x- \frac{9}{4}}$. That tells us that x ranges from 0 to $\frac{9}{4}$ and that, for each x, y ranges from $\frac{3}{2}- \sqrt{x- \frac{9}{4}}$ to $\frac{3}{2}+ \sqrt{x- \frac{9}{4}}$.
i'm sorry but the the limit of 3y-y^2 should be raised to 1/2

4. Okay, correcting for that:

$\int_{y= 0}^3\int_{x= 0}^{\sqrt{3y- y^2}} \sqrt{x^2+ y^2} dxdy$
$x= (3y- y^2)^{1/2}$ leads to $x^2= 3y- y^2$ or $x^2+ y^2- 3y= 0$ so $x^2+ y^2- 3y+ 9/4= 9/4$. That is the same as $x^2+ (y- 3/2)^2= 9/4$, a circle with center at (0, 3/2) and radius 3/2. The original function, then, is the right semicircle.

So when you said "changing the coordinates" you did NOT mean "reverse the order of integration" but rather "change to polar coordinates"!

To stay on the right side of the y-axis, take $\theta$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ and, of course, r from 0 to 3/2. The "differential of area" in polar coordinates is $rdrd\theta$ so your integral becomes $\int_{r=0}^{3/2}\int_{\theta= -\frac{\pi}{2}}^{\frac{\pi}{2}} r^2 d\theta dr$.

That should be easy!