# Thread: changing bounds from rectangular to polar

1. ## changing bounds from rectangular to polar

i just have a general question about how to change the bounds of integration from rectangular to polar. is it as simple as just substituting in y=rsinx (x will represent theta since i don't know how to make theta) and x=rcosx? or do i need to know what the bounds look like and then remake them based on that?

for instance, if a draw a picture of the rectangular bounds, would i then use that picture to help me figure out what the new polar bounds are?

so let's say i have an integral bounded by y <x< root(2-y^2)

would i simply plug in the polar formulas for y, or would i draw out those bounds, look at the picture, and figure out what the proper polar coordinates would be? i'm having trouble doing it either way, as i can't figure out what the polar coordinates would be even after drawing out the picture. and i don't know how to figure out the exact angles that theta is bounded by. if you have any advice, i would greatly appreciate it. i can do most of the polar integration except for these instances when i have to change the bounds to polar.

2. I'm a little unclear about what you're asking. By bounds do you mean region? Are you doing a double integral over some rectangular region and want to convert to polars? If so, the general procedure is to look at the boundary curves for the region and write each one in polars. The limits of integration are determined by the intersection points of the boundary curves.

The example you've given looks incomplete. However, the line $y=x$ with $x>0$ is $\theta =\pi/4$ in polars and the circle $x^2+y^2=2$ is
$r=\sqrt 2$. Limits can't be determined until you specify the rest of the boundary for the region.

3. Geometrically, $x= \sqrt{2- x^2}$ is a semi-circle and y= x is a radius of that semi-circle. The line divides the semi-circle into two regions. The requirement that y< x means we are looking at the lower region. The line y= x corresponds to $\theta= \pi/4$ and the negative y-axis to $\theta= -\pi/2$. And, of course, r goes from 0 to $\sqrt{2}$, the radius of the circle.

4. HallsofIvy,

I don't see where you're getting $\theta =-\pi/2$ from. $y means the region below the line $y=x$. $x<\sqrt{2-y^2}$ potentially allows for negative $x$ values.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}