# Double integrals with polar co-ords

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• Apr 9th 2011, 06:17 AM
olski1
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.
• Apr 9th 2011, 06:24 AM
Prove It
Try converting your region back to Cartesians just so you can get a picture of what the region looks like...
• Apr 9th 2011, 07:44 AM
chisigma
Quote:

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...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Apr 9th 2011, 10:39 PM
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?
• Apr 9th 2011, 11:17 PM
chisigma
Quote:

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 $\displaystyle f(\theta)$ is periodic of period $\displaystyle 2 \pi$ the integral $\displaystyle \displaystyle \int_{\alpha}^{\alpha + 2 \pi} f(\theta)\ d \theta$ is independent from $\displaystyle \alpha$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Apr 10th 2011, 01:29 AM
olski1
Quote:

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

I am not quite sure if i understand what you mean.
• Apr 10th 2011, 01:53 AM
Pranas
Quote:

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