# Bounded region.

• Dec 29th 2008, 06:33 AM
varunnayudu
Bounded region.
Find the area of the region bounded by $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and above the x-axis.
• Dec 29th 2008, 07:00 AM
Last_Singularity
Do you actually need to carry out the integration?

If not, you can simply use the fact that the area of an ellipse is $\displaystyle \pi ab$. And since you are only looking for the area above the x-axis, you just cut that in half to get: $\displaystyle \frac{1}{2} \pi ab$
• Dec 29th 2008, 07:04 AM
Isomorphism
Observe that $\displaystyle \text{Area } = \int_{-a}^{a} y dx = \int_{-a}^{a} \sqrt{b^2 - \frac{b^2 x^2}{a^2}}\, dx$

with the substitution $\displaystyle x = a \sin \theta$,

$\displaystyle \int_{-a}^{a} \sqrt{b^2 - \frac{b^2 x^2}{a^2}}\, dx = \int_{-\frac{\pi}2}^{\frac{\pi}2} b \cos \theta (a \cos \theta ) \, d\theta$$\displaystyle = ab \int_{-\frac{\pi}2}^{\frac{\pi}2} \cos^2 \theta \, d\theta$

Now you can continue :D

Aliter:

Define $\displaystyle x = au$ and $\displaystyle y = bv$. Now we can see that $\displaystyle dx \, dy = ab \, du \, dv$

$\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1$ transforms $\displaystyle u^2 + v^2 = 1$

Thus $\displaystyle \int_{-a}^{a} y \, dx = \int_{-1}^{1} ab\, v \, du = ab$(Area of the semi circle)$\displaystyle = ab(\frac{\pi}2) = \frac{\pi ab}2$
• Dec 29th 2008, 07:13 AM
Soroban
Hello, varunnayudu!

This is a classic problem -- the area of an ellipse (well, semi-ellipse).

Quote:

Find the area of the region bounded by $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} \:= \:1$ and above the x-axis.

The graph is an ellipse, center at (0,0), major axis (ħa,0), minor axis (0, ħb)

Due its symmetry, we can find the area in Quadrant 1 and double.

Solving for $\displaystyle y\!:\;\;y \:=\:\tfrac{b}{a}\sqrt{a^2-x^2}$

The area of the semi-ellipse is: .$\displaystyle A \;=\;2 \times \frac{b}{a}\int^a_0\sqrt{a^2-x^2}\,dx$

Let: $\displaystyle x = a\sin\theta \quad\Rightarrow\quad dx = a\cos\theta\,d\theta$

Substitute: .$\displaystyle A \;=\;\frac{2b}{a}\int a\cos\theta\,(a\cos\theta\,d\theta) \;=\;2ab\int \cos^2\!\theta\,d\theta \;=\;ab\int(1 + \cos2\theta)\,d\theta$

. . . . . . $\displaystyle = \;ab\left(\theta + \tfrac{1}{2}\sin2\theta\right) \;=\;ab\left(\theta + \sin\theta\cos\theta\right)$

Back-substitute: .$\displaystyle \sin\theta = \frac{x}{a},\;\;\cos\theta = \frac{\sqrt{a^2-x^2}}{a}$

. . And we have: .$\displaystyle ab\left(\arcsin\frac{x}{a} + \frac{x\sqrt{a^2-x^2}}{a^2}\right)\,\bigg]^a_0$

Evaluate: .$\displaystyle ab\bigg[(\arcsin1 + 0) - (\arcsin0 + 0)\bigg] \;=\;ab\left(\frac{\pi}{2}\right) \;=\;\frac{1}{2}\pi ab$

• Dec 29th 2008, 07:19 AM
varunnayudu
can u explain me what u have done
Can u explain me what u have done here

with the substitution http://www.mathhelpforum.com/math-he...8458655f-1.gif,

http://www.mathhelpforum.com/math-he...a0dac5f7-1.gif

how did u get $\displaystyle x= asin \theta$
• Dec 29th 2008, 07:38 AM
Isomorphism
Quote:

Originally Posted by varunnayudu
Can u explain me what u have done here

with the substitution http://www.mathhelpforum.com/math-he...8458655f-1.gif,

http://www.mathhelpforum.com/math-he...a0dac5f7-1.gif

how did u get $\displaystyle x= asin \theta$

Perhaps my post was not clear.

Soroban's post explains this step clearly. Let: $\displaystyle x = a \sin \theta \Rightarrow dx = a \cos \theta \, d\theta$. Now substitute $\displaystyle x = a \sin \theta$ and $\displaystyle dx = a \cos \theta \, d\theta$ in the integral. Since x is varying from -a to a, $\displaystyle \theta$ will vary from $\displaystyle -\frac{\pi}2$ to $\displaystyle \frac{\pi}2$. Substitute the limit also.
Now do you see what i have done?

This idea is called integration by substitution. Haven't you seen it before?