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Math Help - Bounded region.

  1. #1
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    Red face Bounded region.

    Find the area of the region bounded by \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 \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
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  3. #3
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    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
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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 \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


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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 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 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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