# Vector calculus - area

• May 5th 2011, 12:34 PM
minifhncc
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 $y \ge 0$, $y \ge -x$ and $x^2+y^2 \le 3\sqrt{x^2+y^2}-3x$.

(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
• May 5th 2011, 12:48 PM
TheEmptySet
Quote:

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 $y \ge 0$, $y \ge -x$ and $x^2+y^2 \le 3\sqrt{x^2+y^2}-3x$.

(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

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

Now since

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

This gives that

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

The area can be calculated as follows

$\int_{0}^{\frac{3\pi}{4}}\int_{0}^{3(1-\cos(\theta))}rdrd\theta$
• May 5th 2011, 12:55 PM
minifhncc
OMG THANK YOU THANK YOU!!! I cannot thank you enough!! Thanks again! I'm so stupid arghhh!