# Area of Region

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• Aug 9th 2012, 04:35 PM
Mrdavid445
Area of Region
What is the area of the region in the first quadrant bounded by the x-axis, the line $\displaystyle y=2x$, and the circles $\displaystyle x^2+y^2=20$ and $\displaystyle x^2+y^2=30$?
• Aug 9th 2012, 07:01 PM
GJA
Re: Area of Region
Hi, Mrdavid445.

The area of a region R in the plane is given by

$\displaystyle \int\int_{R}1\cdot dA$,

where dA is the area element. Since the region we're interested in is a portion of an annular region, it's best to use the polar form for dA, which is

$\displaystyle dA=rdrd\theta$.

What remains are to determine the limits of integration and then compute the double integral.

Does this get everything going in the right direction?

Good luck!
• Aug 9th 2012, 11:01 PM
earboth
Re: Area of Region
Quote:

Originally Posted by Mrdavid445
What is the area of the region in the first quadrant bounded by the x-axis, the line $\displaystyle y=2x$, and the circles $\displaystyle x^2+y^2=20$ and $\displaystyle x^2+y^2=30$?

Since you posted your question in the Geometry forum you probably wanted to use a more geometrical way to solve the problem(?).

1. The area in question is the difference of 2 sectors of 2 concentric circles:

$\displaystyle r_1 = \sqrt{20} \approx 4.472$

$\displaystyle r_2 = \sqrt{30} \approx 5.477$

$\displaystyle |\theta| = \arctan(2) \approx 63.435^\circ$

Draw a sketch!

2. The area a is calculated by:

$\displaystyle a = \frac{|\theta|}{360^\circ} \cdot \pi r_2^2 - \frac{|\theta|}{360^\circ} \cdot \pi r_1^2 =\frac{|\theta|}{360^\circ} \cdot \pi (r_2^2 - r_1^2)$

Plug in the known values.