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Thread: Calculus Problem

  1. #1
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    Calculus Problem

    Please See Attached
    Attached Thumbnails Attached Thumbnails Calculus Problem-calculus.jpg  
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  2. #2
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    You are going to be using the formula
    $\displaystyle V=\int^b_aA(x)dx$
    First see that you are taking the integral from $\displaystyle x=0$ until $\displaystyle x=8$ that is when it intersects the x and y axes.

    Now you need to find $\displaystyle A(x)$:
    Notice that $\displaystyle x+2y=8$ implies $\displaystyle y=-\frac{1}{2}x+8$.
    At $\displaystyle x$ the width of the line perpendicular to the x-axis is then $\displaystyle -\frac{1}{2}x+8$. But this is the diameter of the circle. Thus, its radius is $\displaystyle -\frac{1}{4}x+4$.
    The area of this cross section (semicircle) is
    $\displaystyle \pi \frac{1}{2}\left(-\frac{1}{4}x+4\right)^2$ (remember the 1/2 it is semi-circle )
    Thus,
    $\displaystyle V=\pi \int^8_0\frac{1}{2}\left(-\frac{1}{4}x+4\right)^2dx=\pi \cdot 37.\bar 3\approx 116$
    I think there is something wrong with your choices.
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  3. #3
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    Dear PHckr:

    Line #5 of your narrative reads:


    Notice that implies .

    Do you see the problem?

    Regards,

    Rich B.
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  4. #4
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    Quote Originally Posted by Rich B.
    Dear PHckr:

    Line #5 of your narrative reads:


    Notice that implies .

    Do you see the problem?

    Regards,

    Rich B.
    Sorry
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