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

1. ## 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??

2. Do you mean $\displaystyle \displaystyle A = b\;\int_0^a \sqrt{1-\frac{x^2}{a^2}}\;dx\;?$

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

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

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

so, $\displaystyle \displaystyle A = b\;\int_0^a \sqrt{1-\frac{x^2}{a^2}}\;dx\;=\; \dfrac{b}{a} \int_0^a \sqrt{a^2-x^2}\;dx$

3. I would have done that slightly differently. Let $\displaystyle x= sin(\theta)$ so that $\displaystyle dx= cos(\theta)d\theta$ and $\displaystyle \sqrt{1-\frac{x^2}{a^2}}= \sqrt{1- sin^2(\theta)}= \sqrt{cos^2(\theta)}= cos(\theta)$. When x= 0, $\displaystyle sin(\theta)= 0$ so that $\displaystyle \theta= 0$ and when x= a, $\displaystyle sin(\theta)= 1$ so that $\displaystyle \theta= \pi/2$. The integral becomes
$\displaystyle b\int_0^{\pi/2}cos^2(\theta)d\theta$.

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