HowTo:upper and lowerlimit of integral from Cartesian coordinate to polar coordinate?

**x3bnm** · August 10th 2010, 02:41 PM

I'm stuck in finding integration of a integral. I'm given a integral and asked to change the Cartesian
integral into an equivalent polar integral and then evaluate the resulting integral.

The integral is:
$ \int_0^6\int_0^y\,\, x\,\, dx\,\, dy $

So i replaced $x\,\,dx\,\,dy$ with $r^2\,\, \cos(\theta)\,\,dr\,\,d\theta$ .

The book says:
$ \int_0^6\int_0^y\,\, x\,\, dx\,\, dy = \int_\frac{\pi}{4}^\frac{\pi}{2}\int_0^{6\csc(\the ta)}\,r^2\,\cos(\theta)\,dr\,d\theta $

But how do you find the upper and lower boundary of $r$ and $\theta$ in polar coordinate from Cartesian coordinate?

The book I'm following is vague about the procedure.

There must be a way to find it. But can't seem to figure it out.

Can anyone kindly explain the method?

**yeKciM** · August 10th 2010, 03:52 PM

Originally Posted by x3bnm

I'm stuck in finding integration of a integral. I'm given a integral and asked to change the Cartesian
integral into an equivalent polar integral and then evaluate the resulting integral.

The integral is:
$ \int_0^6\int_0^y\,\, x\,\, dx\,\, dy $

So i replaced $x\,\,dx\,\,dy$ with $r^2\,\, \cos(\theta)\,\,dr\,\,d\theta$ .

The book says:
$ \int_0^6\int_0^y\,\, x\,\, dx\,\, dy = \int_\frac{\pi}{4}^\frac{\pi}{2}\int_0^{6\csc(\the ta)}\,r^2\,\cos(\theta)\,dr\,d\theta $

But how do you find the upper and lower boundary of $r$ and $\theta$ in polar coordinate from Cartesian coordinate?

The book I'm following is vague about the procedure.

There must be a way to find it. But can't seem to figure it out.

Can anyone kindly explain the method?

first of all , when change to polar coordinates, you must know that there are "natural" limits of new (how u wrote) r and $\theta$
but that only means that boundary can't go beyond the "natural" limits, but can go under...

if there is no "restriction" then they will be natural but if u have let's say some conditions then they probably will be changed

$x =\rho \cos \varphi$
$y =\rho \sin \varphi$
$J = \rho$

I prefer to use $\rho$ and $\varphi$ but it's the same

natural limits
$\rho \ge 0$
$\varphi \mid _0 ^{2\pi}$

most times your boundary will be shown with some region, like few lines, parabola, circle's or anything else, and now u put in that conditions your polar x,y, and solve ... to find how does your boundary change

in your example there is 2 boundary's , first is line x=y or y=x , and second y=6... so what that means it means that your integral will look like on the graph

put the "new" x,y, in that equations and see what conditions will u get

$y=x$

$\rho \cos \varphi = \rho \sin \varphi$

$\cos \varphi = \sin \varphi$

$1=\tan \varphi$

$\displaystyle \varphi = \frac {\pi}{4}$

now secon...
$y=6$

can u do it ? and conclude something

try and we'll see are you on the right track

**x3bnm** · August 10th 2010, 07:13 PM

First of all thank you for taking the time and effort for explaining my question. The method you showed works for certain kind of integral. Say we have
$ \int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy $

Now:
$ x = \sqrt{1-y^2}\,\,\,\,\,\,\,\,\, (1) $

$ x = -\sqrt{1-y^2}\,\,\,\,\,\,\,\,\,\ (2) $

$ y = 1\,\,\,\,\,\,\,\,\,\ (3) $

$ y = -1\,\,\,\,\,\,\,\,\,\ (4) $

if we plug in $x =\rho \cos \varphi$ and $y =\rho \sin \varphi$ into equation (1) and solve for $\rho$ we get:
$ \rho = 1 $

Then plug in $y = \rho \sin \varphi$ and $\rho = 1$ into equation (3) and solve for $\varphi$ we get:
$ \varphi = \frac{\pi}{2} $

Again if we plug in $y = \rho \sin \varphi$ and $\rho = 1$ into equation (4) and solve for $\varphi$ we get:
$ \varphi = \frac{3\pi}{2} $

So the upper and lower bound of $\rho$ is $1$ and $0$ respectively and upper and
lower bound for $\varphi$ is $\frac{3\pi}{2}$ and $\frac{\pi}{2}$ respectively.
$ \int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy = \int_{\frac{\pi}{2}}^\frac{3\pi}{2}\int_0^1\,\,r^3 \,\,d\rho\,\,d\varphi $

But the answer is:
$ \int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy = \int_0^{2\pi} \int_0^1\,\,r^3\,\,d\rho\,\,d\varphi $

Why is the upper limit for $\varphi$ is $2\pi$ instead of $\frac{3\pi}{2}$ (what i found) and lower bound 0 instead of
$\frac{\pi}{2}$ (what i found)? Why didn't the method you showed didn't work here? Did i make a mistake here?

Note: I edited this post couple of times to be correct. Sorry about that.

**x3bnm** · August 10th 2010, 10:10 PM

I found the solution to my problem. For those who are interested don't use x boundary equations with y's
boundary equations(don't mix them) to find $r$ and $\theta$ .

So solve two sets of equations separately(so there will be two equations for each x and y) for $r$
and $\theta$ . Again I thank yeKciM for expletive answer.

**yeKciM** · August 10th 2010, 10:18 PM

Originally Posted by x3bnm

First of all thank you for taking the time and effort for explaining my question. The method you showed works for certain kind of integral. Say we have
$ \int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy $

Now:
$ x = \sqrt{1-y^2}\,\,\,\,\,\,\,\,\, (1) $

$ x = -\sqrt{1-y^2}\,\,\,\,\,\,\,\,\,\ (2) $

$ y = 1\,\,\,\,\,\,\,\,\,\ (3) $

$ y = -1\,\,\,\,\,\,\,\,\,\ (4) $

if we plug in $x =\rho \cos \varphi$ and $y =\rho \sin \varphi$ into equation (1) and solve for $\rho$ we get:
$ \rho = 1 $

Then plug in $y = \rho \sin \varphi$ and $\rho = 1$ into equation (3) and solve for $\varphi$ we get:
$ \varphi = \frac{\pi}{2} $

Again if we plug in $y = \rho \sin \varphi$ and $\rho = 1$ into equation (4) and solve for $\varphi$ we get:
$ \varphi = \frac{3\pi}{2} $

So the upper and lower bound of $\rho$ is $1$ and $0$ respectively and upper and
lower bound for $\varphi$ is $\frac{3\pi}{2}$ and $\frac{\pi}{2}$ respectively.
$ \int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy = \int_{\frac{\pi}{2}}^\frac{3\pi}{2}\int_0^1\,\,r^3 \,\,d\rho\,\,d\varphi $

But the answer is:
$ \int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\,\,(x^2+y^2)\,\,dx\,\,dy = \int_0^{2\pi} \int_0^1\,\,r^3\,\,d\rho\,\,d\varphi $

Why is the upper limit for $\varphi$ is $2\pi$ instead of $\frac{3\pi}{2}$ (what i found) and lower bound 0 instead of
$\frac{\pi}{2}$ (what i found)? Why didn't the method you showed didn't work here? Did i make a mistake here?

Note: I edited this post couple of times to be correct. Sorry about that.

wait, u have there circle $x^2+y^2=1$ and how u done is that you look at it like 2 functions, first $\sqrt{1-y^2}$ and second $-\sqrt{1-y^2}$ . Let's look how that looks

(drawing)

u looked at function like that, and you did get that function goes from $\pi /2$ to $3\pi /2$ ... that is true but u (if u look at it like that) have still and another function (right one) which will go from $3\pi/2$ to $\pi/2$ .... that's why you get that

now for that integral you wrote :
$\displaystyle \int _{-1} ^1 dy \int _{-\sqrt{1-y^2}} ^{\sqrt{1-y^2}}(x^2+y^2) \; dx$

hehehehehe

that limit from -1 to 1 you didn't get from equation of line y=1 and y=-1 so that u put and do that what you have done ... that -1 to 1 is from the equation of circle , so that $\varphi$ is on his natural limits from zero to $2\pi$

you don't have any another condition by which your $\varphi$ will change limits

Edit: sorry i didn't see your post, been writing this

**x3bnm** · August 10th 2010, 11:15 PM

That makes a lot of sense. You are absolutely right. Isn't math fun or what? Thanks for extending a helping hand.

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