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Math Help - Area between curve and x axis

  1. #1
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    Area between curve and x axis

    If y = √ 4 - x^2 then the area of the region limited by this curve y and the x axis is

    a)
    b) 2

    Calculate the correct result.
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  2. #2
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    Quote Originally Posted by bret80
    If y = √ 4 - x^2 then the area of the region limited by this curve y and the x axis is

    a)
    b) 2

    Calculate the correct result.
    You have,
    y=\sqrt{4-x^2}
    A semi-circle with radius 2.
    Instead of finding the complicated,
    \int_{-2}^2 \sqrt{4-x^2}dx
    Simplfy, use the formula from geometry.
    The area of the circle is,
    \pi (2)^2=4\pi
    Since this is half the answer is,
    2\pi
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  3. #3
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    Hello, bret80!

    If y \:= \:\sqrt{4 - x^2}, then the area of the region limited by y and the x-axis is

    a)\;\pi\qquad b)\;2\pi

    Calculate the correct result.

    If you are expected to do this with Calculus,
    . . you need to know Trig Substitution (and a bit of Trig).
    I'll baby-step through it for you . . .

    A\;=\;\int^2_{-2}\sqrt{4-x^2}\,dx

    Let x = 2\sin\theta\quad\Rightarrow\quad dx = 2\cos\theta\,d\theta
    . . and \sqrt{4 - x^2} \:= \:\sqrt{4 - 4\sin^2\theta} \:=\:\sqrt{4(1 - \sin^2\theta)} \:=\:\sqrt{4\cos^2\theta} \:=\:2\cos\theta

    Substitute: . A\;=\;\int(2\cos\theta)\,(2\cos\theta\,d\theta) \;= \;4\int\cos^2\theta\,d\theta


    Double-angle identity: . \cos^2\theta \:=\:\frac{1 + \cos2\theta}{2}

    We have: . A\;=\;4\int\left(\frac{1 + \cos2\theta}{2}\right)\,d\theta \;= \;2\int(1 + \cos2\theta)\,d\theta

    Then: . A \;= \;2\left(\theta + \frac{1}{2}\sin2\theta\right) \;= \;2\left(\theta + \sin\theta\cos\theta\right)\,\bigg]^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}} **

    . . A\;= \;2\left[\frac{\pi}{2} + \sin\frac{\pi}{2}\cos\frac{\pi}{2}\right] - 2\left[\text{-}\frac{\pi}{2} + \sin\left(\text{-}\frac{\pi}{2}\right)\cos\left(\text{-}\frac{\pi}{2}\right)\right]

    . . A \;= \;2\left[\frac{\pi}{2} + (1)(0)\right] - 2\left[\text{-}\frac{\pi}{2} + (\text{-}1)(0)\right] \;= \;\pi + \pi \;= \;\boxed{2\pi}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    A change of limits . . .

    Since x \,= \,2\sin\theta\quad\Rightarrow\quad\sin\theta \,= \,\frac{x}{2}

    . . . . \begin{array}{ccc}x \,= \,\text{-}2,\quad\Rightarrow\quad \theta \,= \,\text{-}\frac{\pi}{2} \\ \\x \,= \,2,\quad\Rightarrow\quad\theta \,= \,\frac{\pi}{2}\end{array}

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