# Thread: Double integrals with polar co-ords

1. ## Double integrals with polar co-ords

hello,

I am having trouble deconstructing this problem given to me in class.

i have been asked to sketch the region and evaluate the double integral of

g(r,theta)=1, with the region bounded by r=1-cos(theta).

I dont even really know what the region looks like so i cant begin to set bounds for my double integral.

Any help would be greatly appreciated.

2. Try converting your region back to Cartesians just so you can get a picture of what the region looks like...

3. Originally Posted by olski1
hello,

I am having trouble deconstructing this problem given to me in class.

i have been asked to sketch the region and evaluate the double integral of

g(r,theta)=1, with the region bounded by r=1-cos(theta).

I dont even really know what the region looks like so i cant begin to set bounds for my double integral.

Any help would be greatly appreciated.
The double integral can be written as...

$\displaystyle \int_{-\pi}^{\pi} d \theta\int_{0}^{1-\cos \theta} r\ d r$

... so that it consists in two sucessive integrations...

Kind regards

$\chi$ $\sigma$

4. Okay, so i drew the cardoid described in my question, found my bounds for the inner integral to be from 0 to 1-cos(theta)

but with my outter integral, why is it -pi to pi, does the cardoid not cover a region from 0 to 2pi?

5. Originally Posted by olski1
Okay, so i drew the cardoid described in my question, found my bounds for the inner integral to be from 0 to 1-cos(theta)

but with my outter integral, why is it -pi to pi, does the cardoid not cover a region from 0 to 2pi?
If a function $f(\theta)$ is periodic of period $2 \pi$ the integral $\displaystyle \int_{\alpha}^{\alpha + 2 \pi} f(\theta)\ d \theta$ is independent from $\alpha$...

Kind regards

$\chi$ $\sigma$

6. Originally Posted by chisigma
If a function $f(\theta)$ is periodic of period $2 \pi$ the integral $\displaystyle \int_{\alpha}^{\alpha + 2 \pi} f(\theta)\ d \theta$ is independent from $\alpha$...

Kind regards

$\chi$ $\sigma$
I am not quite sure if i understand what you mean.

7. Originally Posted by olski1
I am not quite sure if i understand what you mean.
Funny, it's exactly the same thing I am trying to prove at this very moment http://www.mathhelpforum.com/math-he...on-177414.html

Don't know if it's enough of a clarification to you though