The length of coil winding on cylinder.

If you imagine unrolling the cylinder the wire is the hypotenuse of a right triangle. The base is ten times around the cylinder and the height is the height of the coil. You can find the length of the wire by Pythagoras. There is no need for integration.


You are working a little harder than you need to. But, you are so close to the finish.

$f(l,r)=$$\int_{0}^{l} \sqrt{(-\frac{20\pi r\sin(\frac{20\pi t}{l})}{l})^2+(\frac{20\pi r\cos(\frac{20\pi t}{l})}{l})^2+(1)^2}dt$

Use the trig identity $\cos^2 t + \sin^2 t = 1$

$\int_{0}^{l} \sqrt{(\frac{20\pi r}{l})^2+(1)^2}dt$

Notice that the integrand doesn't rely on $t.$

$\frac {\sqrt{(20\pi r)^2+l^2}}{l}\int_{0}^{l} dt\\ \sqrt{(20\pi r)^2+l^2}$


From $$r(t)=\begin{bmatrix} r\cos(\frac{20\pi t}{l}) \\ r\sin(\frac{20\pi t}{l}) \\ t \end{bmatrix}$$We get $$ || r'(t) ||= \sqrt {1+r^2(\frac {20\pi }{l}})^2 $$ Therefore the length of the coil is $$ \int_{0}^{l} || r'(t) || dt = \sqrt {\;400r^2\pi^2+l^2}\;$$

Tags:

Integration

Parametrization

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