1. 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!!

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

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

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

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

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

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

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

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

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

Wolfram Mathematica Online Integrator

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

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