# I have to solve this equation or term in order to calculate the area

• February 12th 2011, 03:53 PM
Riazy
I have to solve this equation or term in order to calculate the area
Hi, my teacher did the following step, and I cant understand it really:

We have A = b $(With limits from 0 to a) Sqroot [1-x^2 / a^2 ] dx he got the common denominator as following: a^2 -x^2 / a^2 Now he suddenly wrote it like this .... b/a$ Sqroot [a^2 - x^2] dx My queston is how did he do this step??
• February 12th 2011, 03:59 PM
harish21
Do you mean $\displaystyle A = b\;\int_0^a \sqrt{1-\frac{x^2}{a^2}}\;dx\;?$

For $\sqrt{\dfrac{a^2-x^2}{a^2}}$,

Note: for every a>0, $\sqrt{a^2} = (a^2)^{\frac{1}{2}}= a$

therefore, $\sqrt{\dfrac{a^2-x^2}{a^2}}\;=\;\dfrac{\sqrt{a^2-x^2}}{a}\;=\; \dfrac{1}{a} \times \sqrt{a^2-x^2}$

so, $\displaystyle A = b\;\int_0^a \sqrt{1-\frac{x^2}{a^2}}\;dx\;=\; \dfrac{b}{a} \int_0^a \sqrt{a^2-x^2}\;dx$
• February 13th 2011, 02:33 AM
HallsofIvy
I would have done that slightly differently. Let $x= sin(\theta)$ so that $dx= cos(\theta)d\theta$ and $\sqrt{1-\frac{x^2}{a^2}}= \sqrt{1- sin^2(\theta)}= \sqrt{cos^2(\theta)}= cos(\theta)$. When x= 0, $sin(\theta)= 0$ so that $\theta= 0$ and when x= a, $sin(\theta)= 1$ so that $\theta= \pi/2$. The integral becomes
$b\int_0^{\pi/2}cos^2(\theta)d\theta$.

Or, since this was posted in "Pre-algebra and Algebra", I might recognize that $y= b\sqrt{1- \frac{x^2}{a^2}$ leads to $y^2= b^2\left(1- \frac{x^2}{a^2}\right)$ or $\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$, an ellipse which has are $\pi ab$. Since a square root is always positive, this is the upper portion of that area and has area $\frac{\pi ab}{2}$.