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Thread: Surface area over y-axis (integrating wrong)

  1. #1
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    Surface area over y-axis (integrating wrong)

    The book says the answer is

    $\displaystyle 15\pi\sqrt{17}$

    I have tried to manipulate the numbers to try to get that answer, but I don't know what mistake(s) I am doing.


    Surface area over y-axis (integrating wrong)-surfaceareayaxis.jpg
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  2. #2
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    Re: Surface area over y-axis (integrating wrong)

    Hey Hearts.

    You are using the wrong variable if you are rotating about the y-axis - it should be a function of y not a function of x if you are indeed rotating it about the x-axis.

    The derivative factor will also have to change meaning you will be looking at dx/dy not dy/dx since the slopes will be different.

    If you can state the formula you are using it would be appreciated - otherwise you will have to state how you got it. You are correct in that you will have to take the arc-lengths of each cross section and sum them to get the surface area [which is sort of what the integral is looking to do].

    If you took an arc-length approach you should get a lot of circles with varying cross sections that have their origin at the y-axis and you sum them up. If you do the arc-length approach you can introduce an argument of theta from 0 to 2*pi and make that a function of y and use that to get the area.

    I'll wait for your response.
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    Re: Surface area over y-axis (integrating wrong)

    $\displaystyle \sqrt{1+\left(\frac{1}{4}\right)^2}$

    instead of

    $\displaystyle \sqrt{1+4^2}$
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  4. #4
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    Re: Surface area over y-axis (integrating wrong)

    Quote Originally Posted by chiro View Post
    Hey Hearts.

    You are using the wrong variable if you are rotating about the y-axis - it should be a function of y not a function of x if you are indeed rotating it about the x-axis.

    The derivative factor will also have to change meaning you will be looking at dx/dy not dy/dx since the slopes will be different.

    If you can state the formula you are using it would be appreciated - otherwise you will have to state how you got it. You are correct in that you will have to take the arc-lengths of each cross section and sum them to get the surface area [which is sort of what the integral is looking to do].

    If you took an arc-length approach you should get a lot of circles with varying cross sections that have their origin at the y-axis and you sum them up. If you do the arc-length approach you can introduce an argument of theta from 0 to 2*pi and make that a function of y and use that to get the area.

    I'll wait for your response.
    Quote Originally Posted by Idea View Post
    $\displaystyle \sqrt{1+\left(\frac{1}{4}\right)^2}$

    instead of

    $\displaystyle \sqrt{1+4^2}$
    OHHH now I see what I did wrong. I had dismissed the derivative I should have used, because the wrong one I did gave me sqrt(17) like in the answer from the book. I didn't think to work the WHOLE problem out that way first. I did the 2pi f(x) sqrt(1+f'x^2) formula. It was correct. I just didn't replace the variables everywhere.

    Thank you both very much.
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