Thread: polar coordinates-changing bounds of integration

1. polar coordinates-changing bounds of integration

When I change the integral to polar form, how do i adjust the bounds of integration? Do i have to look at the graph of the new function and determine r and theta from that? or is there a formula to do so? i mean i how to change x and y into rcostheta and rsintheta. but the bounds of integration for theta always involve pi, which i can't get from just some number.

i don't have a specific example, but say the bounds of integration for theta were -3 to 3. is there a formulaic way of figuring out the new bounds, or would i need to look at the graph of the function to determine that?

and when changing the bounds for r, same question basically. is it just using x=rcostheta, etc. or is it using the graph to see where r goes?

thanks for any help

2. Originally Posted by isuckatcalc
When I change the integral to polar form, how do i adjust the bounds of integration? Do i have to look at the graph of the new function and determine r and theta from that? or is there a formula to do so? i mean i how to change x and y into rcostheta and rsintheta. but the bounds of integration for theta always involve pi, which i can't get from just some number.

i don't have a specific example, but say the bounds of integration for theta were -3 to 3. is there a formulaic way of figuring out the new bounds, or would i need to look at the graph of the function to determine that?

and when changing the bounds for r, same question basically. is it just using x=rcostheta, etc. or is it using the graph to see where r goes?

thanks for any help
You will need to post a concrete question that exemplifies what you are asking.

3. You need to make at least a rough sketch of the graph.

The bounds for theta do not have to involve pi. In textbooks they often do because "nice numbers" make the problems seem easier.

After drawing the picture, remember that r is the distance outward from the origin, and theta is the angle made with the positive x-axis.

4. updated with problem

ok. so i have:
sin(x^2+y^2)

y is bounded from 0 to root(9-x^2)
x is bounded from -3 to 3

so changing x^2 and y^2 inside the sin function would leave me with r^2, as the cos^2 and sin^2 go to 1.

so i am now integrating for sin(r^2)? now i am confused, because i don't know what the graph looks like. i can't use my calculator because in polar i can't graph with r as a variable.

also, again, how does this now affect the bounds of integration?
any starting tips would be great, i don't need it to be solved (though feel free )

5. Square both sides of $y= \sqrt{9- x^2}$ to get $y^2= 9- x^2$ or $x^2+ y^2= 9$. Recognize that? It is the circle with center at the origin and radius 3. It intersects the x-axis, nicely enough, at x= -3 and x= 3. Since y is the positive square root your region of integration is the upper semi-circle.

Now you should be able to see what bounds r and $\theta$ have. To integrate $sin(r^2)$, remember that the differential of area, dxdy, becomes $r drd\theta$ in polar coordinates. Use the substitution $u= r^2$.

(What kind of graphing calculator do you have? Most, if not all, modern graphing calculators have a "mode" key that allows you to change to graphing equations in polar coordinates.)

6. awesome, thank so you much. that makes perfect sense. i just have to practice on some other examples now.

i have a ti-89, though i just got it and am far from knowing how to use all of it. i know how to graph polar functions though.

7. a new concern

ok, so i got that one all figured out, and i'm working on another.
it is:
x+y dxdy
x bounded from y to root(2-y^2)
y bounded from 0 to 1

so i converted to polar, and get the double integral of (cos+sin)r^2 drdtheta

i integrate that and get (1/3)r^3(cos+sin) + theta

using the same method as above, i figured the shape is a circle with a radius of root 2

so r is bounded from 0 to root 2

but now how do i find the theta bounds? i have the answer given to me but i can't figure out how to get to it. i made the assumption it was from 0 to pi again, staying on the postive side of the x axis. but that's not getting me the correct answer.

8. I assume you have drawn a picture and recognize that the region to be integrated over is the section of the circle with radius $\sqrt{2}$ and bounded by the lines y= 0 and y= x. At y= 0, $\theta= 0$ of course. The slope of y= x is 1 and the slope is the tangent of the angle. What is that angle? (Another way of looking at it: y= x is exactly half way between x= 0 and y= 0.)