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Math Help - Definite intergral

  1. #1
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    Definite intergral

    Hi,
    I am solving a problem which involves calculating the surface are of a solid revolution about the x – axis. I have found that the surface area can be given by:

    A = 2\Pi \int_{-70}^{110}r+\frac{rd}{\sqrt{r^{2}-x^{2}}}\: dx

    I have done this using the trapezium rule as I know the values of r and d, so I have a rough estimate of a figure I am aiming for.
    The question I am trying to answer says to use the standard integral:

    \int \frac{dx}{\sqrt{k^{2}-x^{2}}}= \tan^{-1}\left ( \frac{x}{\sqrt{k^{2}-x^{2}}} \right )+C

    I am struggling to interpret the r+rd/√(r^2-x^2) into the form dx/√(k^2-x^2) in order to evaluate the integral. Any help or guidance would be appreciated as I have spent too much time trying to do this.

    Thanks
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  2. #2
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    Re: Definite intergral

    \frac{d}{dx}\left (\arcsin{x} \right) = \frac{1}{\sqrt{1-x^2}}

    Assuming r > 0:
    \frac{r}{\sqrt{r^2-x^2}}=\frac{1}{\sqrt{1-(x/r)^2}}

    This is very close to the derivative posted above. What is the derivative of \arcsin(x/r)?
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  3. #3
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    Re: Definite intergral

    \int_{-70}^{110}r+\frac{rd}{\sqrt{r^{2}-x^{2}}}\: dx

    = \int_{-70}^{110}r \: dx+ \int_{-70}^{110}\frac{rd}{\sqrt{r^{2}-x^{2}}}\: dx

    = r\int_{-70}^{110}1 \: dx + rd  \int_{-70}^{110}\frac{1}{\sqrt{r^{2}-x^{2}}}\: dx

    Is this what you mean?
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  4. #4
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    Re: Definite intergral

    Hello, freethinker!

    I am solving a problem which involves calculating the surface area of a solid revolution about the x–axis.

    I have found that the surface area can be given by: . A \:=\:2\pi \int_{-70}^{110}\left(r+\frac{rd}{\sqrt{r^{2}-x^{2}}}\right)\,dx

    I have done this using the trapezium rule as I know the values of r and d,
    so I have a rough estimate of a figure I am aiming for.

    The question says to use the standard integral: . \int \frac{dx}{\sqrt{k^{2}-x^{2}}}\:=\: \tan^{-1}\!\left ( \frac{x}{\sqrt{k^{2}-x^{2}}} \right )+C
    . . This is NOT a standard formula!

    I am struggling to interpret the r+\frac{rd}{\sqrt{r^2-x^2}} into the form \frac{dx}{\sqrt{k^2-x^2}} in order to evaluate the integral.
    . . Really? . . . What's stopping you?

    Any help or guidance would be appreciated as I have spent too much time trying to do this. Thanks.

    That "standard integral" would be: . \int\frac{dx}{\sqrt{r^2-x^2}} \:=\:\sin^{\text{-}1}\!\left(\frac{x}{r}\right)+ C


    We have: . 2\pi r \int^{110}_{\text{-}70}\left(1 + \frac{d}{\sqrt{r^2-x^2}}\right)\,dx \;=\;2\pi r\left[x + d\sin^{-1}\!\left(\frac{x}{r}\right)\right]^{110}_{\text{-}70}

    Now finish it . . .

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  5. #5
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    Re: Definite intergral

    Thank you for your help, I have finished the problem now

    I was just having a basic algebra problem, and that standard integral information being incorrect threw me a bit
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