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Math Help - Volume using Washer/Disc method

  1. #1
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    Volume using Washer/Disc method

    I have a problem I'm working on for a study guide, but it's worded a little oddly compared to the other problems.

    Set up the integral and find the volume of the bounded region with the washer/disc method.

    15. Below by y=3x^2_1 and above by y=4, rotated about the line y=4.

    It doesn't seem like it should be that hard, but again the wording is throwing me off. Any help would be much appreciated. The answer is (48pi/5). I just need to figure out how to get there.
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    Re: Volume using Washer/Disc method

    Quote Originally Posted by rberg1897 View Post
    I have a problem I'm working on for a study guide, but it's worded a little oddly compared to the other problems.

    Set up the integral and find the volume of the bounded region with the washer/disc method.

    15. Below by y=3x^2_1 and above by y=4, rotated about the line y=4.

    It doesn't seem like it should be that hard, but again the wording is throwing me off. Any help would be much appreciated. The answer is (48pi/5). I just need to figure out how to get there.
    I think something is wrong with your question. Here is a graph of the region in question.

    Volume using Washer/Disc method-graph.jpg

    It will have infinite volume if it is rotated around y=4
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    Re: Volume using Washer/Disc method

    I think he means the bounded part between the parabola and the line (bounded below by the parabola and bounded above by the line). If so, the integral goes like:

    \int_{min(x)}^{max(x)} \pi (radius)^2 dx

    where radius is 4 - (3x^2-1), and min(x) and max(x) are the x-coordinates of the intersection points.

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