# Arc length & area of surface of revolution

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• Jun 4th 2012, 02:01 AM
hiy312
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(Happy)
• Jun 4th 2012, 06:06 AM
Reckoner
Re: Arc length & area of surface of revolution
Quote:

Originally Posted by hiy312
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 $\displaystyle k=\pi\Rightarrow u=2$.

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

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

$\displaystyle =\frac{128\pi}5$
• Jun 4th 2012, 12:23 PM
Reckoner
Re: Arc length & area of surface of revolution
Quote:

Originally Posted by hiy312
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.
• Jun 4th 2012, 10:47 PM
hiy312
Re: Arc length & area of surface of revolution
Really pro! Thanks so much Reckoner!

Quote:

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

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

But I don't understand this step, why will cos (k-2k) turn into cos k?
• Jun 4th 2012, 10:49 PM
hiy312
Re: Arc length & area of surface of revolution
Quote:

Originally Posted by Reckoner
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!
• Jun 5th 2012, 04:32 AM
Reckoner
Re: Arc length & area of surface of revolution
Quote:

Originally Posted by hiy312
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.
• Jun 5th 2012, 05:37 AM
Reckoner
Re: Arc length & area of surface of revolution
Quote:

Originally Posted by hiy312
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.
• Jun 6th 2012, 12:37 AM
hiy312
Re: Arc length & area of surface of revolution
Really helped me a lot! A million thanks to you Reckoner!!