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Math Help - Help with area and volume problems

  1. #1
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    Help with area and volume problems

    Could someone please help me with the following area and volume problems?

    -Find the volume of the solid formed when the region bounded by the curves y=x^3 +1, x=1, and y=0 is rotated about the x-axis. How do I do something like this for a volume of the solid problem with three curves???

    -Find the area bounded by the curves f(x)=x^3 +x^2 and g(x)=2x^2 +2x.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Missylovesmathz View Post
    Could someone please help me with the following area and volume problems?

    -Find the volume of the solid formed when the region bounded by the curves y=x^3 +1, x=1, and y=0 is rotated about the x-axis. How do I do something like this for a volume of the solid problem with three curves???
    I believe you will use the disc method to solve. Multiple curves are needed to define an area to be rotated. A particular curve can determine the integrand, or a limit of integration. Maybe try looking at an example in your textbook? It helps to draw pictures.

    Quote Originally Posted by Missylovesmathz View Post
    -Find the area bounded by the curves f(x)=x^3 +x^2 and g(x)=2x^2 +2x.
    We need to know on what interval.
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Missylovesmathz View Post
    Could someone please help me with the following area and volume problems?

    -Find the volume of the solid formed when the region bounded by the curves y=x^3 +1, x=1, and y=0 is rotated about the x-axis. How do I do something like this for a volume of the solid problem with three curves???

    -Find the area bounded by the curves f(x)=x^3 +x^2 and g(x)=2x^2 +2x.



    second

    x^3 + x^2 = 2x^2+2x \Rightarrow x^3-x^2-2x = 0 \Rightarrow x=\; 2,\; 0\; -1

    f(\frac{-1}{2}) > g(\frac{-1}{2})
    f(1) < g(1) so the bounded area

    \int_{-1}^{0} f(x) - g(x) \; dx + \int_{0}^{2} g(x) - f(x) \; dx
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  4. #4
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    What the problem states, is that you need to find the solid rotation between y=x^3 +1, x=1, and the x-axis y=0.

    How y=0 does help you? You don't even need to draw the graphic.

    Just find x^3+1=0, that is the intersection between the x-axis and y=x^3+1.

    In problems like these usually you need to find intersection points, and it is often very useful to draw the graphics of the functions.

    So you need to find the solid rotation of x^3+1 from -1 to 1. Integrals will help you do that.

    @undefined you do not need boundaries. The graphs of the functions are intersecting. Just find the intersecting points.

    Regards.
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  5. #5
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Amer View Post
    second

    x^3 + x^2 = 2x^2+2x \Rightarrow x^3-x^2-2x = 0 \Rightarrow x=\; 2,\; 0\; -1

    f(\frac{-1}{2}) > g(\frac{-1}{2})
    f(1) < g(1) so the bounded area

    \int_{-1}^{0} f(x) - g(x) \; dx + \int_{0}^{2} g(x) - f(x) \; dx
    Ah you're right, I was a bit lazy and graphed it, but the zoom was too wide and I hastily jumped to a wrong conclusion about how the curves intersected. Thanks.

    Quote Originally Posted by p0oint View Post
    @undefined you do not need boundaries. The graphs of the functions are intersecting. Just find the intersecting points.

    Regards.
    Yes yes, I noticed, just wasn't fast enough typing my correction. Thanks.
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