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Thread: Bounded region.

  1. #1
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    Red face 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.
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  2. #2
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    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: $\displaystyle \frac{1}{2} \pi ab$
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  3. #3
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    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

    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$
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  4. #4
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    Hello, varunnayudu!

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


    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 $


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  5. #5
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    can u explain me what u have done

    Can u explain me what u have done here

    with the substitution ,




    how did u get $\displaystyle x= asin \theta$
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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by varunnayudu View Post
    Can u explain me what u have done here

    with the substitution ,




    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?
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