# integration to find area please

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• December 5th 2009, 01:12 PM
pepsi
integration to find area please
hi

I want to find the area inside the circle r=2, with x>0 and y>1

I evaluate the

integral (from x=0 to x=2) of the ingral (y=1 to y=+sqrd.root.of(4-x^2)of dydx

and come up with pi-2 = 1.14 but book answer is 1.23

pls help!!
• December 5th 2009, 01:32 PM
adkinsjr
Quote:

Originally Posted by pepsi
hi

I want to find the area inside the circle r=2, with x>0 and y>1

I evaluate the

integral (from x=0 to x=2) of the ingral (y=1 to y=+sqrd.root.of(4-x^2)of dydx

and come up with pi-2 = 1.14 but book answer is 1.23

pls help!!

The integral $\int_0^2\sqrt{4-x^2}dx$ is the area of $\frac{1}{4}$ of the circle. So the total area should be:

$4\int_0^2\sqrt{4-x^2}dx$

The indefinite integral is $I(x)=\frac{1}{2}\sqrt{4-x^2}x+2sin^{-1}\left(\frac{x}{2}\right)$

http://integrals.wolfram.com/index.j...D&random=false

So did you calculate $4[I(2)-I(0)=4\pi$

Your book must be wrong. The area of a circle with radius $2$ is $4\pi$. You can show this using integration as I did above.
• December 5th 2009, 01:50 PM
pepsi
Quote:

Originally Posted by adkinsjr
The integral $\int_0^2\sqrt{4-x^2}dx$ is the area of $\frac{1}{4}$ of the circle. So the total area should be:

$4\int_0^2\sqrt{4-x^2}dx$

The indefinite integral is $I(x)=\frac{1}{2}\sqrt{4-x^2}x+2sin^{-1}\left(\frac{x}{2}\right)$

Wolfram Mathematica Online Integrator

So did you calculate $4[I(2)-I(0)=4\pi$

Your book must be wrong. The area of a circle with radius $2$ is $4\pi$. You can show this using integration as I did above.

thanks for your reply,

however the question is not to find the area of the circle but that which is defined by x>0 and y>1 and the cirlce, that is, a subsection of the first orthant...