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given

$$r=4e^{3\theta} \space \space \space \space dr/d\theta=(3*4*e^{3\theta})$$ $$l=\int \sqrt(4e^{3\theta})^{2}+(3*4*e^{3\theta})^{2} \rightarrow $$

why does the integral $$ \int\sqrt(16e^{6\theta}+144e^{6\theta}) \rightarrow \int\sqrt(160e^{6\theta})$$ become $$ 4\sqrt(10) \int\sqrt(e^{6\theta} \rightarrow somehow \rightarrow (4/3) \sqrt(10) (e^{6 \pi}-1)$$ instead of $$ 4\sqrt(10) \int \sqrt(e^{12\theta} \rightarrow U=\sqrt(e^{12})\space ; \space DU=e^{12\theta}*12)$$ $$ \rightarrow(4 \sqrt(10)*[((2/3)e^{12 \theta}]*[e^{12 \theta}*12]\rightarrow (8/3) \sqrt(10) (e^{12 \theta}*12) du\space went \space away$$ I'm confused about the integration of e after I move the constants to the other side of the integral. I don't have the steps so I can only guess that there was a 1/3 somewhere in the integration of the first equation. this in an equation for length of a polar curve. I have omitted the original function r was equal too.

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    $\begingroup$ The very second line baffles me....what does that mean? $\endgroup$
    – DonAntonio
    Dec 7, 2013 at 0:14
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    $\begingroup$ I'm already baffled by the first line. $\endgroup$
    – Daniel Fischer
    Dec 7, 2013 at 0:14
  • $\begingroup$ Hehe...by starting a sentence without a capital letter, @DanielFischer ? $\endgroup$
    – DonAntonio
    Dec 7, 2013 at 0:16
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    $\begingroup$ @DonAntonio That is the zeroth line. (whistles innocently) $\endgroup$
    – Daniel Fischer
    Dec 7, 2013 at 0:17
  • $\begingroup$ sorry I had an int in the wrong place $\endgroup$
    – user113979
    Dec 7, 2013 at 0:32

1 Answer 1

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You have not included it from your post, but from your answer I believe that your questions is to evaluate the length of the curve $r = 4e^{3\theta}$ for $0 \leq \theta \leq 2\pi$.

The formula for arc length is: $$L = \int_a^{b} \sqrt{r^2 + (\frac{dr}{d\theta})^2} \,d\theta$$

Then we have $\frac{dr}{d\theta} = 12e^{3\theta}$, so substituting in we get:

$$L = \int_0^{2\pi} \sqrt{160e^{6\theta}} \,d\theta = \sqrt{160}\int_0^{2\pi} \sqrt{e^{6\theta}} \,d\theta = \sqrt{160}\int_0^{2\pi} e^{3\theta} \,d\theta$$.

Then using the substitution $u = 3\theta$, we find that $$L = \sqrt{160}\int_0^{2\pi} e^{3\theta} \,d\theta = \frac{\sqrt{160}}{3}\int_0^{6\pi} e^{u} \,du = \frac{4\sqrt{10}}{3}(e^{6\pi}-1)$$

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  • $\begingroup$ so you added r and dr before applying the square? which is why you have $$160e^{6\theta}$$ and not $$160e^{12\theta}$$ or did I do something wrong with the exponets $\endgroup$
    – user113979
    Dec 7, 2013 at 1:10
  • $\begingroup$ By exponent rules, we know that $(e^a)^b = e^{ab}$. Thus $(e^{3\theta})^2 = e^{6\theta}$. When you add exponential functions, you do not add the powers (that is true when you multiply them). So $16e^{6\theta}+ 144e^{6\theta} = 160e^{6\theta}$. $\endgroup$
    – Tim
    Dec 7, 2013 at 1:13
  • $\begingroup$ yeah that's what I thought. BTW in the last step when you n integrate u is it $$e^{(u^{2}/2)}=e^{((3\pi)^{2}/2)}=(e^{6\pi}-1)$$ $\endgroup$
    – user113979
    Dec 7, 2013 at 1:25
  • $\begingroup$ Recall that the antiderivative of $e^u$ is just $e^u$ (As the derivative of $e^u$ is $e^u$). What I did was make a u-substitution and changed the limits of integration by plugging the original bounds into our u-substitution to get new bounds. The point of this is to avoid having to change the $u$'s back to $\theta$'s when we plug in our bounds. Thus, $\int_0^{6\pi} e^u \,du = e^u |_{0}^{6\pi} = e^{6\pi} - 1$. $\endgroup$
    – Tim
    Dec 7, 2013 at 1:29
  • $\begingroup$ oh,yeah that's right I missed the bounds. I usually always switch back. $\endgroup$
    – user113979
    Dec 7, 2013 at 1:48

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