3
$\begingroup$

I'm having trouble with this question on my Calculus III homework. I'm not really sure how to go about this one honestly.

A lamina is bounded by the curve $r = 2\cos(3θ)\space$ when $-π/6 <= θ <= π/6$. Assuming a constant density, $ρ = k$, give the rectangular coordinates of the centroid.

For clarity of the question, and clarity of what type of answer I am looking for:

My professor taught me the following: $$M_x = \iint y\rho(x,y)dA$$ $$M_y = \iint x\rho(x,y)dA$$ $$m = \iint \rho(x,y)dA$$ The answer I need is, $$(\bar x, \bar y) = \left(\frac{M_y}{m},\frac{M_x}{m}\right)$$ In addition, I was taught that $$ x = r\cos(\theta) $$ $$ y = r\sin(\theta) $$ when converting between polar and rectangular coordinates.

Any assistance would be greatly appreciated. Thanks.

$\endgroup$
  • $\begingroup$ You need to find centre of mass, right ? $\endgroup$ – A---B Apr 17 '17 at 15:17
  • $\begingroup$ Yes. @A---B that is correct. $\endgroup$ – Tommy Yamakaitis Apr 17 '17 at 15:20
  • $\begingroup$ The solution that you accepted below is incorrect. Please see my answer below that. $\endgroup$ – Cye Waldman Apr 28 '17 at 16:04
2
$\begingroup$

Let's first calculate the total mass of the centroid. The total mass $M$ is given by

$$\begin{align} M&=\int_{-\pi/6}^{\pi/6} \int_0^{2\cos(3\theta)}\rho \,r\,dr\,d\theta\\\\ &=4\rho\int_0^{\pi/6}\cos^2(3\theta)\,d\theta\\\\ &=\frac\pi3\rho \end{align}$$

Because of the symmetry of the centroid, we expect the center of mass to lie on the $x$-axis. In fact, the center of mass, $\vec r_{CM}=\hat x \bar x+\hat y \bar y$, is given by

$$\begin{align} \vec r_{CM}&=\frac1M\int_{-\pi/6}^{\pi/6}\int_0^{2\cos(3\theta)} \vec r\,\rho\,r\,dr\,d\theta\\\\ &=\frac3\pi\left(\hat x \int_{-\pi/6}^{\pi/6}\int_0^{2\cos(3\theta)} r^2\cos(\theta)\,dr\,d\theta+\hat y \underbrace{\int_{-\pi/6}^{\pi/6}\int_0^{2\cos(3\theta)} r^2\sin(\theta)\,dr\,d\theta}_{=0\,\text{due to odd symmetry}}\right)\\\\ &=\hat x \frac{16}{\pi}\int_0^{\pi/6}\cos^3(3\theta)\cos(\theta)\,d\theta \tag 1 \end{align}$$

The evaluation of the integral on the right-hand side of $(1)$ is left as an exercise.

$\endgroup$
  • $\begingroup$ I'm not sure how you got 4p/3. Also, I need the rectangular coordinates for the center of mass. $\endgroup$ – Tommy Yamakaitis Apr 17 '17 at 16:43
  • $\begingroup$ This is rectangular coordinates. What notation do you not understand? $\endgroup$ – Mark Viola Apr 17 '17 at 16:44
  • $\begingroup$ Okay, thank you. Now, what is ${\hat x}$? $\endgroup$ – Tommy Yamakaitis Apr 17 '17 at 17:09
  • $\begingroup$ $\hat x$ is a unit vector along the $x$ axis. $\endgroup$ – Mark Viola Apr 17 '17 at 17:10
  • $\begingroup$ This wasn't exactly what I was looking for, but it did help me understand. I think I get it now, thank you! $\endgroup$ – Tommy Yamakaitis Apr 17 '17 at 17:24
2
$\begingroup$

Let me demonstrate how much simpler this problem is in the complex plane. First of all, we note that curve can be expressed as

$$z=2\cos(3\theta)e^{i\theta}$$

The arc length, area, and centroid are given by

$$ s=\int |\dot z| d\theta\\ A=\frac{1}{2}\int \mathfrak{Im}\{z^*\dot z\}d\theta\\ z_c=\frac{1}{3A}\int z\ \mathfrak{Im}\{z^*\dot z\}d\theta $$

Thus, we need $\dot z$, $z^*$, $\mathfrak{Im}\{z^*\dot z\}$ as follows

$$\dot z=[3\cdot 2\sin(3\theta)+i\cdot2\cos(3\theta)]e^{i\theta}\\ z^*=2\cos(3\theta)e^{-i\theta}\\ \mathfrak{Im}\{z^*\dot z\}=4\cos^2(3\theta) $$

Then we can readily show that

$$ A=\frac{1}{2}\int_{-\pi/6}^{\pi/6}4\cos^2(3\theta)d\theta=\frac{2}{3}\int_{-\pi/2}^{\pi/2}\cos^2\varphi d\varphi=\frac{\pi}{3}\\ z_c=\frac{8}{3A}\int_{-\pi/6}^{\pi/6}\cos^3(3\theta)e^{i\theta}d\theta=\frac{8}{3A}\int_{-\pi/6}^{\pi/6}\cos^3(3\theta)\cos\theta d\theta=\frac{8}{\pi}\left(\frac{81\sqrt{3}}{320}+0i\right) $$

Notice the cosine arrangement in the last integral.

$\endgroup$
  • $\begingroup$ How is this an improvement over a very straightforward real analysis approach? In fact, it obfuscates the simplicity, I believe. Please explain your claim "how much simpler this problem is in the complex plane." $\endgroup$ – Mark Viola Apr 29 '17 at 1:25
  • $\begingroup$ @Dr.MV Thank you for your comment. I wrote an essay in response and it was 942 characters over. So this will have to suffice for now. The complex variable contains more information than the corresponding Cartesian or polar variables. As such, many relations can be simplified. In the present problem, the area and centroid have been expressed as single integrals, instead of doubles. Those integrals are readily solved numerically exactly as is in the complex plane. In fact, I verified my solutions numerically; it took exactly 5 lines of code to get the area and centroid. $\endgroup$ – Cye Waldman Apr 29 '17 at 16:19
  • $\begingroup$ Cye, I don't believe that your assertion is valid. Look at the solution that I posted and observe its simplicity and efficiency. $\endgroup$ – Mark Viola Apr 29 '17 at 17:41
  • $\begingroup$ @Dr.MV Mark, we'll just have to agree to disagree. We must likely feel more comfortable with what we know best. Suppose, the curve in question was given parametrically? I would handle it in exactly the same way. In addition, there are many things that can be done in the complex plane that simply cannot be done in Cartesian or polar coordinates. Case in point: show me an equation, say $y(x)$ or $r(\theta)$ for the Fibonacci or Padovan spiral. $\endgroup$ – Cye Waldman Apr 29 '17 at 18:31
  • $\begingroup$ Cye, that's fine. To clarify the "what we know best" comment, my PhD is in Electromagnetic Field Theory, a discipline involving $3$-D vector fields. Analysis is often facilitated using integral transforms (e.g., Fourier Transforms) that result in vector fields with complex-valued components. So, I am perfectly comfortable with complex analysis. That said, the straightforward calculation of the center of mass is not, I believe, facilitated by complex analysis, despite its elegance as you presented herein. So, as you wrote, we will agree to disagree. Hopefully, we are MSE friends. -Mark $\endgroup$ – Mark Viola Apr 29 '17 at 19:12
0
$\begingroup$

EDIT1

$$ {\bar x} = \frac{\int r ^3\cos \theta d\theta}{\int r^2 d\theta},\, {\bar y} = \frac{\int r ^3\sin \theta d\theta}{\int r^2 d\theta} $$

Plugin $r$ and integrate between given $\theta$ limits. The latter vanishes as for an odd function.

$\endgroup$
  • $\begingroup$ Could you please elaborate? How will this give me the rectangular coordinates? @Narasimham $\endgroup$ – Tommy Yamakaitis Apr 17 '17 at 15:24
  • $\begingroup$ Okay, so my question is, would I plugin $2\cos(3θ)$ for $r$ in these instances? and $-π/6$ and $π/6$ for $θ$? @Narasimham $\endgroup$ – Tommy Yamakaitis Apr 17 '17 at 15:41
  • $\begingroup$ This is not correct. The position vector $\vec r$ is not given by $\hat r r+\hat \theta \theta$. $\endgroup$ – Mark Viola Apr 17 '17 at 15:47
  • $\begingroup$ @ Tommy Yamakaitis Apologies. I put in in more direct form in edit that I actually wanted to say $\endgroup$ – Narasimham Apr 17 '17 at 16:00
  • $\begingroup$ So, you're also saying ${\bar y} = 0$? $\endgroup$ – Tommy Yamakaitis Apr 17 '17 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.