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I've been giving a good time trying to solve this problem, I do not find a clear way to solve appreciate your help. \begin{array}{rcl} r& =& \sqrt{\cos 2\theta } \end{array} This Around to axis y and the limits of integration is 0 to π/4

P.d : I would appreciate if anyone could recommend me some text or page in which you could learn to plot in polar and learn more about these problems.

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  • $\begingroup$ Plotting Lemniscate of Bernoulli the figure is defined for $ \theta = \pm \pi/4 $ in polar coordinates. $\endgroup$ – Narasimham May 15 '15 at 15:05
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Parametric Breakdown

Start by dividing the function from polar form into a set of two parametric functions. That is, the horizontal and vertical component of the radial coordinate $r$:

$$ x =r \cos{\theta}\qquad\quad y=r\sin{\theta}\\ \implies \dfrac{dx}{d\theta}=\dfrac{dr}{d\theta}\cos{\theta}-r\sin{\theta} \qquad\quad \dfrac{dy}{d\theta} = \dfrac{dr}{d\theta}\sin{\theta}+r\cos{\theta}\\ $$

However the components alone don't really help, we need to find a differential element $ds$ that represents the hypotenuse formed by the two differential components (basically Pythagorean Theorem) :

\begin{align} (ds)^2 &= (\dfrac{dx}{d\theta})^2+(\dfrac{dy}{d\theta})^2\\ & = (\dfrac{dr}{d\theta}\cos{\theta}-r\sin{\theta})^2 + (\dfrac{dr}{d\theta}\sin{\theta}+r\cos{\theta})^2\\ & = \left[(\dfrac{dr}{d\theta})^2\cos^2\theta\color{red}{-(\dfrac{dr}{d\theta})2r\cos\theta\sin\theta} +r^2\sin^2\theta\right] \space\space\space + \left[(\dfrac{dr}{d\theta})^2\sin^2\theta \color{red}{+ (\dfrac{dr}{d\theta})2r\cos\theta\sin\theta} +r^2cos^2\theta \right]\\ & =(\dfrac{dr}{d\theta})^2(\cos^2\theta + \sin^2\theta) + r^2(\cos^2\theta + \sin^2\theta)\\ &=(\dfrac{dr}{d\theta})^2+r^2 \end{align}

Hence we conclude for this step that:

$$ds = \sqrt{(\dfrac{dr}{d\theta})^2+r^2} d\theta$$

Surface of Revolution Derivation

We should notice (through derivation using frustrums not shown here) that just as in moving from the arc length of a curve to its surface area in parametric coordinates requires multiplication by $2\pi * \left(x(t) \,\mathrm{or}\,y(t)\right)$, the same applies in polar coordinates:

$$ L = \int ds $$

Since we are rotating about the y-axis, the height of each frustrum would be the $x$ component of the polar equation:

$$A = 2\pi\int \operatorname{x}(\theta)\, ds\\ \boxed{A = 2\pi\int \operatorname{x}(\theta)\,\sqrt{(\dfrac{dr}{d\theta})^2+r^2} d\theta} $$

If you wish to try it from here yourself, don't continue reading as I will propose the calculated solution!

Calculated Solution

Recollecting variables:

$r = \sqrt{\cos2\theta}\\ \operatorname{x}(\theta)= r\cos\theta = \cos\theta\sqrt{\cos2\theta}\\ \dfrac{dr}{d\theta}= \dfrac{-2\sin2\theta}{2\sqrt{\cos2\theta}} = \dfrac{-\sin2\theta}{\sqrt{\cos2\theta}} $

Plugging in variables and simplifying:

\begin{align} A &= 2\pi\int_0^{\pi/4} \operatorname{x}(\theta)\,\sqrt{(\dfrac{dr}{d\theta})^2+r^2}\, d\theta\\ & = 2\pi\int_0^{\pi/4}\cos\theta\sqrt{\cos2\theta}\,\sqrt{(\dfrac{-\sin2\theta}{\sqrt{\cos2\theta}})^2+(\sqrt{\cos2\theta})^2}\, d\theta\\ & = 2\pi\int_0^{\pi/4}\cos\theta\sqrt{\cos2\theta}\,\sqrt{\dfrac{\sin^22\theta}{\cos2\theta}+\cos2\theta}\, d\theta \end{align}

Combining the square roots and using the fact that $\sin^2x+\cos^2x=1$:

\begin{align} &= 2\pi\int_0^{\pi/4}\cos\theta\,\sqrt{\cos2\theta \left(\dfrac{\sin^22\theta}{\cos2\theta}+\cos2\theta\right)}\, d\theta\\ &= 2\pi\int_0^{\pi/4}\cos\theta\,\sqrt{\sin^22\theta + \cos^22\theta}\, d\theta\\ &= 2\pi\int_0^{\pi/4}\cos\theta\,\sqrt{(1)}\, d\theta\\ &= 2\pi\, \left.\sin\theta\right\rvert_0^{\pi/4}\\ &= 2\pi\, \left(\dfrac{\sqrt{2}}{2}\right)\\ & = \boxed{\pi\sqrt{2}} \end{align}

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  • $\begingroup$ Thanks bro i very apreciate.. $\endgroup$ – user239295 May 21 '15 at 0:59
  • $\begingroup$ @user239295 No problem :) $\endgroup$ – Hushus46 May 21 '15 at 2:46

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