# Area between curve and x axis

• Aug 3rd 2006, 10:09 AM
bret80
Area between curve and x axis
If y = √ 4 - x^2 then the area of the region limited by this curve y and the x axis is

a) ¶
b) 2¶

Calculate the correct result.
• Aug 3rd 2006, 10:10 AM
ThePerfectHacker
Quote:

Originally Posted by bret80
If y = √ 4 - x^2 then the area of the region limited by this curve y and the x axis is

a) ¶
b) 2¶

Calculate the correct result.

You have,
$\displaystyle y=\sqrt{4-x^2}$
$\displaystyle \int_{-2}^2 \sqrt{4-x^2}dx$
Simplfy, use the formula from geometry.
The area of the circle is,
$\displaystyle \pi (2)^2=4\pi$
Since this is half the answer is,
$\displaystyle 2\pi$
• Aug 3rd 2006, 01:08 PM
Soroban
Hello, bret80!

Quote:

If $\displaystyle y \:= \:\sqrt{4 - x^2}$, then the area of the region limited by $\displaystyle y$ and the x-axis is

$\displaystyle a)\;\pi\qquad b)\;2\pi$

Calculate the correct result.

If you are expected to do this with Calculus,
. . you need to know Trig Substitution (and a bit of Trig).
I'll baby-step through it for you . . .

$\displaystyle A\;=\;\int^2_{-2}\sqrt{4-x^2}\,dx$

Let $\displaystyle x = 2\sin\theta\quad\Rightarrow\quad dx = 2\cos\theta\,d\theta$
. . and $\displaystyle \sqrt{4 - x^2} \:=$ $\displaystyle \:\sqrt{4 - 4\sin^2\theta} \:=\:\sqrt{4(1 - \sin^2\theta)} \:=\:\sqrt{4\cos^2\theta} \:=\:2\cos\theta$

Substitute: .$\displaystyle A\;=\;\int(2\cos\theta)\,(2\cos\theta\,d\theta) \;= \;4\int\cos^2\theta\,d\theta$

Double-angle identity: .$\displaystyle \cos^2\theta \:=\:\frac{1 + \cos2\theta}{2}$

We have: .$\displaystyle A\;=\;4\int\left(\frac{1 + \cos2\theta}{2}\right)\,d\theta \;= \;2\int(1 + \cos2\theta)\,d\theta$

Then: .$\displaystyle A \;= \;2\left(\theta + \frac{1}{2}\sin2\theta\right) \;= \;2\left(\theta + \sin\theta\cos\theta\right)\,\bigg]^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}}$ **

. . $\displaystyle A\;= \;2\left[\frac{\pi}{2} + \sin\frac{\pi}{2}\cos\frac{\pi}{2}\right] - 2\left[\text{-}\frac{\pi}{2} + \sin\left(\text{-}\frac{\pi}{2}\right)\cos\left(\text{-}\frac{\pi}{2}\right)\right]$

. . $\displaystyle A \;= \;2\left[\frac{\pi}{2} + (1)(0)\right] - 2\left[\text{-}\frac{\pi}{2} + (\text{-}1)(0)\right] \;= \;\pi + \pi \;= \;\boxed{2\pi}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

A change of limits . . .

Since $\displaystyle x \,= \,2\sin\theta\quad\Rightarrow\quad\sin\theta \,= \,\frac{x}{2}$

. . . . $\displaystyle \begin{array}{ccc}x \,= \,\text{-}2,\quad\Rightarrow\quad \theta \,= \,\text{-}\frac{\pi}{2} \\ \\x \,= \,2,\quad\Rightarrow\quad\theta \,= \,\frac{\pi}{2}\end{array}$