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I have to calculate this integral:

$$\int_Lyds$$ where L is a part of ellipse $$\begin{cases} x=2\cos(t) \\ y=3\sin(t) & \ \end{cases}$$ in first quadrant.

The problem is the integrand contains only $y$ instead of $xy$. That's why "$u$ substitution" doesn't work, since it gives $du\over dt$ equals multiplication of $\sin(t)$ and $\cos(t)$ times some constant, but I have only $3\sin(t)$ to use as a part of my $du$.

Tyberius
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Martins2018
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2 Answers2

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hint

By Pythagoras,

$$ds=\sqrt {dx^2+dy^2}=$$

$$\sqrt {4\sin^2 (t)+9\cos^2 (t)}dt=$$ $$\sqrt{4+5\cos^2 (t)}dt .$$

the integral becomes

$$3\int_0^{\frac {\pi}{2}}\sqrt{4+5\cos^2 (t)}\sin (t)dt=$$

$$6\int_0^1\sqrt {1+\frac {5u^2}{4}}du $$

Now put $u=\frac {2\sinh (v)}{\sqrt {5}} $ and finish it.

hamam_Abdallah
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  • Thanks for the answer. Seems that I have to get familiar with hyperbolic functions since I had no idea that I have to use them here. And doesn't $sinh(v)$ has to contain $t$ instead of $v$? – Martins2018 Mar 03 '18 at 20:08
  • @Martins2018 The variable is dummy. you can replace $v $ by $\theta $ ... – hamam_Abdallah Mar 03 '18 at 20:10
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Hint:

remember that

$$ ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt $$

Emilio Novati
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