Question

Hi
How can I change the Cartesian integral into an polar integral?
Thanks in advance!

Answer 1

I have an exampel her:
\[\int\limits_{0}^{2}\int\limits_{0}^{\sqrt{2x-x^2}}\sqrt{x^2+y^2} dy dx\]

Answer 2

In general,
plug in
\(x = r\cos \theta \\ y = r\sin \theta\)

Answer 3

here,
\(r^2= x^2+y^2\)

Answer 4

find what upper limit 2x-x^2 equals

Answer 5

\(dy dx = r dr d\theta \)

Answer 6

But how can I know that the upper limit 2x-x^2 is in polar?

Answer 7

y = 0 to
y = \(\sqrt{2x-x^2}\) << this is a circle.

can you plot it?

Answer 8

you mean that i should convert the upper limit to polar, right?

Answer 9

Yes I just have plot it now, and it's a half circel

Answer 10

yes,

\(0 \le x \le 2 \\
0\le y \le \sqrt{2x-x^2}\)

is equivalent to
\(
0 \le \theta \le \pi/2 \\ 0\le r\le 2 \)

makes sense?

Answer 11

wait, my limits for 'r' aren't accurate

Answer 12

since we have the circle,
\(x^2+y^2 -2x= 0 \\ (x-1)^2 +y^2 = 1^2\)

we can take
\(x-1 = r \cos \theta \\ y = r\sin \theta\)

Answer 13

\(0 \le x \le 2 \\
0\le y \le \sqrt{2x-x^2} \)
is now equivalent to
\(0\le\ \theta \le \pi\\ 0\le r \le 1 \)

Answer 14

you'll have to now find the value of
\(\sqrt{x^2+y^2}\)

Answer 15

\(\sqrt{(r\cos \theta +1)^2+(r\sin\theta)^2 } = \)

Answer 16

that is r when we take x = r cos and y = r sin ...
this is not what we have taken here

Answer 17

\(\int_{0}^{\pi} \int_0^1 \sqrt{r^2+2r\cos\theta+1}r dr d\theta \)

Answer 18

the conversion is done,
do you also have to evaluate it?

Answer 19

because evaluating that would be a challenge in polar form

Answer 20

Yes it shoudl also be evaluated :/
I use mathcad an maple.. but it can solve it :/

Answer 21

ok, lets try
\(u = r^2+ 2r \cos \theta +1 \\ du = (2r +2\cos \theta) dr \)

Answer 22

There are two competing choices for which polar coordinates to use, since we have the function centered at the origin and the upper bound centered at x=1. It turns out the one centered at the origin is easier, and we'll see in a sec why that's true:

\[x=r \cos \theta\]\[y=r \sin \theta\]
We plug these things into our set of bounds from the integral, \(y=0\), \(y=\sqrt{2x-x^2}\), \(x=0\), \(x=2\). Notice that we can rearrange that upper bound on y to look like this with minor algebra:
\[x^2+y^2 = 2x\]

Anywho, once we plug in these polar coordinates this becomes
\[r^2 = 2 r \cos \theta\]
in other words,
\[r=2 \cos \theta\]

Now if we pullback to this we have:

\[\int_0^2 \int_0^{\sqrt{2x-x^2}} \sqrt{x^2+y^2} dydx = \int_0^\pi \int_0^{2 \cos \theta} r *rdrd\theta \]
\[\int_0^\pi \int_0^{2 \cos \theta} r *rdrd\theta = \int_0^\pi \frac{1}{3} r^3 |_0^{2 \cos \theta} d \theta=\frac{8}{3} \int_0^\pi \cos^3 \theta d \theta\]

Answer 23

nice!
but wouldn't the limit be o to pi/2 here?

Answer 24

the \(\int_0^\pi \cos^3 \theta = 0 \)

Answer 25

must be a typo...
\(\Large \int_0^{\pi/2} \int_0^{2 \cos \theta} r *rdrd\theta \\ \Large = \int_0^{\pi/2} \dfrac{1}{3} r^3 |_0^{2 \cos \theta} d \theta=\dfrac{8}{3} \int_0^{\pi/2} \cos^3 \theta d \theta = \dfrac{16}{3}\)

Answer 26

yeah haha

Answer 27

Really thanks to both of you because you help me!!
Sorry i don't say it but I have the result, It should give 16/9 = 1,78

Answer 28

oh yes,
the cos integral is 2/3
so the final answer would be 16/9 indeed

Answer 29

Just out of curiosity, I wonder if the substitution \(u=x-1\) is worth looking at. It would turn the integral into (probably, I'm sloppy):

\[\int_{-1}^1 \int_0^{\sqrt{1-u^2}} \sqrt{(u+1)^2+y^2} dy du\]

Answer 30

Anywho, I guess I'm just trying to help convey what sorta things are useful to look at in double and triple integrals. It's usually about choosing some choice of coordinates that leverages the symmetry of the situation. Usually you can find this out by taking all of the limits, writing them out and/or drawing them and then looking for where things are important.

In this case we had two different interesting points and it looked like polar was better than rectangular at one of them. But overall there were 4 pretty decent possibilities. (Well, 3 since I guess the original one sucked which is why this question was asked in the first place and we needed to change lol)

Answer 31

Sorry, but I don't get it.. What should I change to 2/3 ?
And sorry becuase of my bad english!

Answer 32

\(\cos 3x = 4 \cos^3 x - 3\cos x\)
\(\int_0^{\pi/2} \cos^3 x dx= \int_0^{\pi/2} \dfrac{\cos 3x + 3\cos x}{4}dx = (1/4)[\dfrac{\sin 3x}{3}+3\sin x ]_0^{\pi/2} \\= (1/4)[-1/3+3-0] = 2/3\)

^^

Answer 33

Can you help me to write a recipe / a step by step how to do list to solve or change cratesian integral into polar integral? :/