# Thread: Vector calculus - area

1. ## Vector calculus - area

Hello. I have no clue how to do this one. I don't know what the region looks like. I've tried squaring etc but with no avail.

"A plane region is determined by $\displaystyle y \ge 0$, $\displaystyle y \ge -x$ and $\displaystyle x^2+y^2 \le 3\sqrt{x^2+y^2}-3x$.

(i) sketch this region
(ii) find its area"

How would I do this?

I've tried this for literally 1.5 hrs and I haven't got it

Thanks

2. Originally Posted by minifhncc
Hello. I have no clue how to do this one. I don't know what the region looks like. I've tried squaring etc but with no avail.

"A plane region is determined by $\displaystyle y \ge 0$, $\displaystyle y \ge -x$ and $\displaystyle x^2+y^2 \le 3\sqrt{x^2+y^2}-3x$.

(i) sketch this region
(ii) find its area"

How would I do this?

I've tried this for literally 1.5 hrs and I haven't got it

Thanks
Hopefully you know polar coordinates if you change you will get

$\displaystyle r^2 \le 3r-3r\cos(\theta) \iff r \le 3(1-\cos(\theta))$

Now since

$\displaystyle y \ge 0 \text{ and } y \ge -x$

This gives that

$\displaystyle \theta \in [0, \frac{3\pi}{4}]$

The area can be calculated as follows

$\displaystyle \int_{0}^{\frac{3\pi}{4}}\int_{0}^{3(1-\cos(\theta))}rdrd\theta$

3. OMG THANK YOU THANK YOU!!! I cannot thank you enough!! Thanks again! I'm so stupid arghhh!