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Math Help - Volume

  1. #1
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    Volume

    I'm having a lot of problems with these questions.

    Volume of the region enclosed by y^2=4x and y=x about the line x=4

    I can graph it, but I can't set up the integral.

    Another one is "The base of a solid is a the region enclosed between the graphs of y=sinx and y=-sinx from x=0 to x=pi. Each cross section perpendicular to the x-axis is a semicircle with diameter connecting the two graphs. Find the volume."

    I just don't know what to do with the information it gives me in the second sentence.
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  2. #2
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    Quote Originally Posted by CarDoor View Post
    I'm having a lot of problems with these questions.

    Volume of the region enclosed by y^2=4x and y=x about the line x=4

    I can graph it, but I can't set up the integral.

    Another one is "The base of a solid is a the region enclosed between the graphs of y=sinx and y=-sinx from x=0 to x=pi. Each cross section perpendicular to the x-axis is a semicircle with diameter connecting the two graphs. Find the volume."

    I just don't know what to do with the information it gives me in the second sentence.
    washers w/r to y about the line x = 4 ...

    V = \pi \int_0^4 \left(4- \frac{y^2}{4}\right)^2 - (4-y)^2 \, dy

    using cylindrical shells w/r to x about x = 4 ...

    V = 2\pi \int_0^4 (4-x)(2\sqrt{x} - x) \, dx<br />



    for the second problem, the diameter of a semicircular cross section is the vertical distance from y = \sin{x} to y = -\sin{x} , therefore, d = 2\sin{x} ... so, the radius of a semi-circular cross section is r = \sin{x}

    cross-sectional area = \frac{\pi}{2}r^2 = \frac{\pi}{2}\sin^2{x}

    V = \int_0^{\pi} \frac{\pi}{2}\sin^2{x} \, dx
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  3. #3
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    "The base of a solid is a the region enclosed between the graphs of y=sinx and y=-sinx from x=0 to x=pi. Each cross section perpendicular to the x-axis is a semicircle with diameter connecting the two graphs. Find the volume."

    The region between the functions can be denoted by

    y=sin(x)-(-sin(x))=2sin(x)

    This is the diameter of the semicircle. Can you picture them stacked up in

    the region with the base of the semicircles being between sin(x) and -sin(x)?.

    The area of a semicircle is \frac{{\pi}y^{2}}{2}

    The radius is \frac{2sin(x)}{2}=sin(x)

    \frac{\pi}{2}\int_{0}^{\pi}sin^{2}(x)dx
    Last edited by galactus; January 12th 2010 at 04:19 PM. Reason: sorry, skeeter. Ya' beat me to it. But hey, we agree:)
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  4. #4
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    Thanks for the help.

    I still have a question though. Why was is 4-y but not 4+y? By looking at the graphing how can you tell?
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  5. #5
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    Quote Originally Posted by CarDoor View Post
    Thanks for the help.

    I still have a question though. Why was is 4-y but not 4+y? By looking at the graphing how can you tell?
    what is the horizontal distance from the line x = 4 to x = y ?
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  6. #6
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    Alright. I think I got it. Thanks again for the help.
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  7. #7
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    Okay. Maybe I don't get it. What if the region was on the right of x=4? Would it be y-4?
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  8. #8
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    Quote Originally Posted by CarDoor View Post
    Okay. Maybe I don't get it. What if the region was on the right of x=4? Would it be y-4?
    right curve - left curve
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  9. #9
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    Alright thanks. I think I'm finally getting comfortable with this.
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