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Math Help - Area between 2 curves.

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    Area between 2 curves.

    Find the area of the region R, enclosed by:

    x+2y=0 and x+y^2=3
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  2. #2
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    Re: Area between 2 curves.

    Quote Originally Posted by softstyll View Post
    Find the area of the region R, enclosed by:

    x+2y=0 and x+y^2=3
    Start by graphing the regions, so that you get an idea of what the region at least looks like.

    Then find out where they intersect, so that you can work out the limits of your integration...
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    Re: Area between 2 curves.

    Quote Originally Posted by Prove It View Post
    Start by graphing the regions, so that you get an idea of what the region at least looks like.

    Then find out where they intersect, so that you can work out the limits of your integration...
    Yep, I did that, the thing is the parabola extends past one of its intersection points with the line. I'm not sure if this creates a problem. Would I have to split the integral into multiple parts?
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    Re: Area between 2 curves.

    Quote Originally Posted by softstyll View Post
    Yep, I did that, the thing is the parabola extends past one of its intersection points with the line. I'm not sure if this creates a problem. Would I have to split the integral into multiple parts?
    You would probably need to use a double integral...
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    Re: Area between 2 curves.

    I don't think you'll need to do multiple parts here.

    Try \displaystyle \int_{a}^b\sqrt{3-x}-\frac{-x}{2}~dx where a and b are the solutions for intersection.
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    Re: Area between 2 curves.

    Have you considered how arbitrary your choice of orientation or variable?

    \int_{-1}^{3}What?\;dy
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    Re: Area between 2 curves.

    Quote Originally Posted by softstyll View Post
    Find the area of the region R, enclosed by:

    x+2y=0 and x+y^2=3
    First let's see where they intersect...

    Subtracting the second equation from the first gives

    \displaystyle \begin{align*} x + y^2 - (x + 2y) &= 3 - 0 \\ y^2 - 2y &= 3 \\ y^2 - 2y - 3 &= 0 \\ (y + 1)(y - 3) &= 0 \\ y &= -1 \textrm{ or }y = 3 \end{align*}

    and the corresponding \displaystyle \begin{align*} x \end{align*} values are \displaystyle \begin{align*}  2\end{align*} and \displaystyle \begin{align*} -6 \end{align*} respectively.

    Now, if we decide to use horizontal strips, they are bounded on the left by the line \displaystyle \begin{align*}  x = -2y \end{align*} and they are bounded on the right by the curve \displaystyle \begin{align*} x = 3 - y^2 \end{align*} . When we sum all these strips, they are bounded below by \displaystyle \begin{align*} y = -1 \end{align*} and bounded above by \displaystyle \begin{align*} y = 3 \end{align*} .

    So the double integral is

    \displaystyle \begin{align*} \int_{-1}^3{\int_{-2y}^{3 - y^2}{\,dx}\,dy} &= \int_{-1}^3{\left[x\right]_{-2y}^{3-y^2}\,dy} \\ &= \int_{-1}^3{\left(3 - y^2\right) - (-2y)\,dy} \\ &= \int_{-1}^3{3 + 2y - y^2\,dy} \\ &= \left[3y + y^2 - \frac{y^3}{3}\right]_{-1}^3 \\ &= \left[3(3) + 3^2 - \frac{3^3}{3}\right] - \left[3(-1) + (-1)^2 - \frac{(-1)^3}{3}\right] \\ &= \left(9 + 9 - 9\right) - \left(-3 + 1 + \frac{1}{3}\right) \\ &= 9 - \left(-\frac{5}{3}\right) \\ &= \frac{32}{3} \end{align*}


    Now why not try using vertical strips to see if you get the same answer?
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