# Integrating using polar co-ordinates.

• Apr 25th 2009, 08:59 AM
Showcase_22
Integrating using polar co-ordinates.
I ran into some trouble with some integrals that use polar coordinates.

Quote:

Change the following cartesian integrals into an equivalent polar integral. Then evaluate the polar integral.
Sounds fair!(Clapping)

Quote:

1). $\int_0^6 \int_0^y x~dx~dy$
....(Angry)

The region for integration is a triangle which I thought was a little weird. I think the r limits start from 0 to something (I called it b in the integral below), but I think the $\theta$ limits are $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$.

So, without the final r limit, I have $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_0^b r^2cos(\theta)~dr~d \theta$.

Quote:

2). $\int_0^2 \int_0^{\sqrt{1-(x-1)^2}}\frac{x+y}{x^2+y^2}~dy~dx$
My problem with this one is that the region of integration is not a circle centred at the origin but rather the point $(1,0)$. I am aware that the region is a semi circle. For this reason, are the $\theta$ limits $0 \leq \theta \leq \pi$?

Are the r limits $0 \leq r \leq \sqrt{1-(rcos(\theta)-1)^2}$?

However, the integrand is going to be $\frac{r(rcos(\theta)+rsin(\theta))}{r^2}=cos(\thet a)+sin(\theta)$.

So overall, I get that the integral should be $\int_0^{\sqrt{1-(rcos(\theta)-1)^2}} \int_0^{\pi} cos(\theta)+sin(\theta)~dr~d \theta$.

Clearly the r limits are incorrect!

From here, I decided to put $\theta=\pi$ into the upper limit of the double integral:

$\int_0^{\sqrt{r^2+2r}} \int_0^{\pi} cos(\theta)+sin(\theta)~dr~d \theta=\int_0^{\pi} \int_0^{\sqrt{r^2+2r}} cos(\theta)+sin(\theta)~d \theta dr$.

Is this integral the correct one to evaluate?
• Apr 25th 2009, 09:14 AM
Jester
Quote:

Originally Posted by Showcase_22
I ran into some trouble with some integrals that use polar coordinates.

Sounds fair!(Clapping)

....(Angry)

The region for integration is a triangle which I thought was a little weird. I think the r limits start from 0 to something (I called it b in the integral below), but I think the $\theta$ limits are $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$.

So, without the final r limit, I have $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_0^b r^2cos(\theta)~dr~d \theta$.

My problem with this one is that the region of integration is not a circle centred at the origin but rather the point $(1,0)$. I am aware that the region is a semi circle. For this reason, are the $\theta$ limits $0 \leq \theta \leq \pi$?

Are the r limits $0 \leq r \leq \sqrt{1-(rcos(\theta)-1)^2}$?

However, the integrand is going to be $\frac{r(rcos(\theta)+rsin(\theta))}{r^2}=cos(\thet a)+sin(\theta)$.

So overall, I get that the integral should be $\int_0^{\sqrt{1-(rcos(\theta)-1)^2}} \int_0^{\pi} cos(\theta)+sin(\theta)~dr~d \theta$.

Clearly the r limits are incorrect!

From here, I decided to put $\theta=\pi$ into the upper limit of the double integral:

$\int_0^{\sqrt{r^2+2r}} \int_0^{\pi} cos(\theta)+sin(\theta)~dr~d \theta=\int_0^{\pi} \int_0^{\sqrt{r^2+2r}} cos(\theta)+sin(\theta)~d \theta dr$.

Is this integral the correct one to evaluate?

Problem 1, $r$ will go from the origin to the line $y = 6$ so $r = 0$ to $r \sin \theta = 6$ or $r = 6 \csc \theta$

Problem 2. It's best to draw a picture of the region. It a half circle (upper half) centered at $x =1$ with radius $r=1$. The polar equation for this is $r = 2 \cos \theta$. To sweep our the area $\theta = 0 \to \pi/2$ and so you have

$
\int_0^{\pi/2} \int_0^{2 \cos \theta} \sin \theta + \cos \theta dr d\theta
$