Bounded region.

**varunnayudu** · December 29th 2008, 06:33 AM

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

**Last_Singularity** · December 29th 2008, 07:00 AM

Do you actually need to carry out the integration?

If not, you can simply use the fact that the area of an ellipse is $\pi ab$ . And since you are only looking for the area above the x-axis, you just cut that in half to get: $\frac{1}{2} \pi ab$

**Isomorphism** · December 29th 2008, 07:04 AM

Observe that $\text{Area } = \int_{-a}^{a} y dx = \int_{-a}^{a} \sqrt{b^2 - \frac{b^2 x^2}{a^2}}\, dx$

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

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

Now you can continue

Aliter:

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

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

Thus $\int_{-a}^{a} y \, dx = \int_{-1}^{1} ab\, v \, du = ab$ (Area of the semi circle) $= ab(\frac{\pi}2) = \frac{\pi ab}2$

**Soroban** · December 29th 2008, 07:13 AM

Hello, varunnayudu!

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

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

The graph is an ellipse, center at (0,0), major axis (±a,0), minor axis (0, ±b)

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

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

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

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

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

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

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

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

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

**varunnayudu** · December 29th 2008, 07:19 AM

Can u explain me what u have done here

with the substitution

,

how did u get $x= asin \theta$

**Isomorphism** · December 29th 2008, 07:38 AM

Originally Posted by varunnayudu

Can u explain me what u have done here

with the substitution

,

how did u get $x= asin \theta$

Perhaps my post was not clear.

Soroban's post explains this step clearly. Let: $x = a \sin \theta \Rightarrow dx = a \cos \theta \, d\theta$ . Now substitute $x = a \sin \theta$ and $dx = a \cos \theta \, d\theta$ in the integral. Since x is varying from -a to a, $\theta$ will vary from $-\frac{\pi}2$ to $\frac{\pi}2$ . 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?

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can u explain me what u have done

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