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Thread: Arc length & area of surface of revolution

  1. #1
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    Arc length & area of surface of revolution

    Hi everyone! I have two questions, one about area of surface of revolution and another is about arc length...
    I really fail to do this two question despite many times of trying so I hope someone can help me

    1. Find the area of the surface of revolution generated by revolving the arc of the cardioid " x = 2 cos k - cos 2k, y = 2 sin k - sin 2k " about the X-axis.
    2. A warehouse is 75m long and 40m wide. A cross-section of the roof is the inverted catenary y = 31 - 10 (e^0.05x + e^-0.05x). Find the number of square metres of roofing in the warehouse. Hint: Find the arc length of the catenary and multiply this by the length of the warehouse.

    I would be really grateful if you can help me
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    Re: Arc length & area of surface of revolution

    Quote Originally Posted by hiy312 View Post
    1. Find the area of the surface of revolution generated by revolving the arc of the cardioid " x = 2 cos k - cos 2k, y = 2 sin k - sin 2k " about the X-axis.
    $\displaystyle S = 2\pi\int_0^\pi\left(2\sin k - \sin2k\right)\sqrt{\left(-2\sin k+2\sin2k\right)^2+\left(2\cos k-2\cos2k\right)^2}\,dk$

    $\displaystyle = 2\pi\int_0^\pi\left(2\sin k - \sin2k\right)\sqrt{\left(4\sin^2k-8\sin k\sin2k+4\sin^22k\right)+\left(4\cos^2k-8\cos k\cos2k+4\cos^22k\right)}\,dk$

    $\displaystyle = 2\pi\int_0^\pi\left(2\sin k - \sin2k\right)\sqrt{8-8\sin k\sin2k-8\cos k\cos2k}\,dk$

    $\displaystyle = 4\pi\sqrt2\int_0^\pi\left(2\sin k - \sin2k\right)\sqrt{1-\left(\cos k\cos2k+\sin k\sin2k\right)}\,dk$

    $\displaystyle = 4\pi\sqrt2\int_0^\pi\left(2\sin k - \sin2k\right)\sqrt{1-\cos(k-2k)}\,dk$

    $\displaystyle = 8\pi\sqrt2\int_0^\pi\left(\sin k - \sin k\cos k\right)\sqrt{1-\cos k}\,dk$

    $\displaystyle = 8\pi\sqrt2\int_0^\pi\left(1 - \cos k\right)\sqrt{1-\cos k}\sin k\,dk$

    $\displaystyle = 8\pi\sqrt2\int_0^\pi\left(1 - \cos k\right)^{3/2}\sin k\,dk$

    Let $\displaystyle u = 1-\cos k\Rightarrow du=\sin k\,dk$. Then $\displaystyle k = 0\Rightarrow u=0$ and $\displaystyle k=\pi\Rightarrow u=2$.

    $\displaystyle S = 8\pi\sqrt2\int_0^2u^{3/2}\,du$

    $\displaystyle =8\pi\sqrt2\left[\frac25u^{5/2}\right]_0^2$

    $\displaystyle =\frac{128\pi}5$
    Thanks from hiy312
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    Re: Arc length & area of surface of revolution

    Quote Originally Posted by hiy312 View Post
    2. A warehouse is 75m long and 40m wide. A cross-section of the roof is the inverted catenary y = 31 - 10 (e^0.05x + e^-0.05x). Find the number of square metres of roofing in the warehouse. Hint: Find the arc length of the catenary and multiply this by the length of the warehouse.
    Our catenary is

    $\displaystyle y = 31 - 10\left(e^{0.05x}+e^{-0.05x}\right)$

    $\displaystyle = 31 - 20\cosh0.05x$,

    so

    $\displaystyle y' = -\sinh0.05x$.

    Assuming that $\displaystyle x$ and $\displaystyle y$ are in meters, the length of the cross section is

    $\displaystyle s = \int_{-20}^{20}\sqrt{1+\sinh^20.05x}\,dx$

    $\displaystyle = \int_{-20}^{20}\cosh0.05x\,dx$

    $\displaystyle =20\sinh0.05x\bigg]_{-20}^{20}$

    $\displaystyle =20\left(\sinh1-\sinh(-1)\right)$

    $\displaystyle =40\sinh1 = \frac{20\left(e^2 - 1\right)}e$

    Now multiply by the length of the warehouse.
    Thanks from hiy312
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    Re: Arc length & area of surface of revolution

    Really pro! Thanks so much Reckoner!

    Quote Originally Posted by Reckoner View Post
    $\displaystyle = 4\pi\sqrt2\int_0^\pi\left(2\sin k - \sin2k\right)\sqrt{1-\cos(k-2k)}\,dk$

    $\displaystyle = 8\pi\sqrt2\int_0^\pi\left(\sin k - \sin k\cos k\right)\sqrt{1-\cos k}\,dk$
    But I don't understand this step, why will cos (k-2k) turn into cos k?
    Last edited by hiy312; Jun 4th 2012 at 10:51 PM.
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    Re: Arc length & area of surface of revolution

    Quote Originally Posted by Reckoner View Post
    Our catenary is

    $\displaystyle y = 31 - 10\left(e^{0.05x}+e^{-0.05x}\right)$

    $\displaystyle = 31 - 20\cosh0.05x$,
    I don't understand it either... why did 10(e^0.05x + e^-0.05x) turn into 20cos h 0.05x?
    Once again, thanks so much!
    Last edited by hiy312; Jun 4th 2012 at 10:52 PM.
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    Re: Arc length & area of surface of revolution

    Quote Originally Posted by hiy312 View Post
    But I don't understand this step, why will cos (k-2k) turn into cos k?
    Remember that cosine is an even function, so $\displaystyle \cos(-x) = \cos x$:

    $\displaystyle \cos(k-2k)=\cos(-k)=\cos k$

    Your other question I can address later when I have more time. For now, note that I used hyperbolic functions to make the integration smoother. You could leave it in exponential form though.
    Thanks from hiy312
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    Re: Arc length & area of surface of revolution

    Quote Originally Posted by hiy312 View Post
    I don't understand it either... why did 10(e^0.05x + e^-0.05x) turn into 20cos h 0.05x?
    Okay, leaving it in exponential form,

    $\displaystyle y = 31 - 10\left(e^{0.05x}+e^{-0.05x}\right)$

    $\displaystyle \Rightarrow y' = -10\left(0.05e^{0.05x}-0.05e^{-0.05x}\right)$

    $\displaystyle =-\frac12\left(e^{0.05x}-e^{-0.05x}\right)$

    Therefore, our arc length is

    $\displaystyle s = \int_{-20}^{20}\sqrt{1+\frac14\left(e^{0.05x}-e^{-0.05x}\right)^2}\,dx$

    $\displaystyle = \int_{-20}^{20}\sqrt{1+\frac14\left(e^{x/10}+e^{-x/10}-2\right)}\,dx$

    $\displaystyle =\frac12\int_{-20}^{20}\sqrt{2+e^{x/10}+e^{-x/10}}\,dx$

    $\displaystyle =\frac12\int_{-20}^{20}\sqrt{\left(e^{x/20}+e^{-x/20}\right)^2}\,dx$

    $\displaystyle =\frac12\int_{-20}^{20}\left(e^{x/20}+e^{-x/20}\right)\,dx$

    $\displaystyle =\frac12\bigg[20e^{x/20}-20e^{-x/20}\bigg]_{-20}^{20}$

    $\displaystyle =10\left(e - e^{-1} - e^{-1} + e\right)$

    $\displaystyle =20\left(e-\frac1e\right)$

    $\displaystyle =\frac{20\left(e^2-1\right)}e$

    And note that this matches our previous result.
    Thanks from hiy312
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    Re: Arc length & area of surface of revolution

    Really helped me a lot! A million thanks to you Reckoner!!
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