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Math Help - Find the area enclosed by the curve and the y-axis

  1. #1
    Junior Member Shadow236's Avatar
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    Find the area enclosed by the curve and the y-axis

    I have to find the area enclosed by the curve: x = t2-2t, y = t1/2 and the y-axis. I know that this is some kind of intergral, but I don't really know how to approach it because it is on the y-axis.

    How would I setup the integral?

    Thank you.
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  2. #2
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    Re: Find the area enclosed by the curve and the y-axis

    Solving for t in terms of x would be ugly, because t shows up twice in that equation. Rather, solve for it in terms of y using the second equation.

    if y = t^{\frac{1}{2}}

    then

    y^2 = t

    x = y^4 - 2y^2 = y^2(y^2-2)

    This is convenient for us because have an expression for x in terms of y, perfect for finding the area between this and the y axis.

    A = |\int_{0}^{\sqrt{2}} y^4 - 2y^2 dy|
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  3. #3
    Junior Member Shadow236's Avatar
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    Re: Find the area enclosed by the curve and the y-axis

    Question: Why is the limit from 0 to sqrt(2)?

    I ended up with area = (4sqrt(2))/5) - (4sqrt(2)/3) -- Is that correct?
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    Junior Member FelixFelicis28's Avatar
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    Re: Find the area enclosed by the curve and the y-axis

    Quote Originally Posted by Shadow236 View Post
    Question: Why is the limit from 0 to sqrt(2)?

    I ended up with area = (4sqrt(2))/5) - (4sqrt(2)/3) -- Is that correct?
    The limit is from 0 to \sqrt{2} because those are the points at which your graph intersects the y-axis - think about it, it intercepts the y-axis when x=0 \implies y^2(y^2 -2) = 0 \implies y = 0, \ \sqrt{2} \big(we discard y = - \sqrt{2} because y is defined by y = \sqrt{t} \implies y > 0 \ \forall t \in \mathbb{R} \big)

    Yes but remember, area is a positive quantity:

    \text{Area} = \left| \int_{0}^{\sqrt{2}}  y^2 (y^2 - 2) \ dy \right|
    Last edited by FelixFelicis28; September 4th 2013 at 06:02 AM.
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  5. #5
    Junior Member FelixFelicis28's Avatar
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    Re: Find the area enclosed by the curve and the y-axis

    Quote Originally Posted by FelixFelicis28 View Post
    \big( y > 0 \ \forall t \in \mathbb{R} \big)
    Scratch that, meant y \geq 0 \ \forall \ t \geq 0
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