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Let S be the area of the region bounded by the curves $$y=\frac 3x,\>\>\> y=\frac 5x,\>\>\> y=3x,\>\>\> y=6x$$ Need to find $S$.

The coordinates of the vertices of the resulting figure were found. The problem with the transition from a double integral to a repeated one.

There's another idea. Make a replacement \begin{cases} \xi=xy \\ \eta=\frac{y}{x} \end{cases} Вut none led to the correct answer.

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  • $\begingroup$ Why not show your work with your replacement variables $\xi,\eta$? Maybe you made a mistake we can spot. $\endgroup$ – MPW Dec 5 '19 at 2:11
  • $\begingroup$ Hi, the change of variables you suggested should work. Can you show more of what you did there? $\endgroup$ – arkeet Dec 5 '19 at 2:12
  • $\begingroup$ hi! my answer is $ln4$. I can give more details later $\endgroup$ – math_14 Dec 5 '19 at 2:45
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The required bounded area by four curves is $$A=\int_{1/\sqrt{2}}^{\sqrt{5/6}}( 6x-3/x) dx+ \int_{\sqrt{5/6}}^{1}~(5/x-3/x) dx+\int_{1}^{\sqrt{5/3}} (5/x-3x) dx $$ $$\implies A=\left(1-\frac{3}{4} \log(5/3) \right)+[\log(6/5)]+ \left( -1+\frac{5}{2} \log (5/3)\right)=\log 2.$$

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In polar coordinates, the boundaries are

$$r^2=\frac3{\sin\theta\cos\theta},\>\>\>\>\>r^2=\frac5{\sin\theta\cos\theta}, \>\>\>\>\>\tan\theta = 3,\>\>\>\>\>\tan\theta = 6$$

Thus, the area integral is,

$$S=\int_{\theta_1}^{\theta_2}d\theta \int_{r_1}^{r_2}rdr =\int_{\theta_1}^{\theta_2}\frac{d\theta}{\sin\theta\cos\theta}=\ln(\tan\theta)|_{\tan^{-1}3}^{\tan^{-1}6}=\ln2$$

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1) Subtract the two integrals of $5/x$ and $3/x$ in the range of $a$ which is the solution of $\frac{3}{x}=6x$ (the intersection of two lines) to $b$ which is the solution of $\frac{5}{x}=3x$.

2) One would obtain $area=[ln(2x)]_{a}^{b}$.

3)Now, to subtract the area of the two hyperbolic triangles, simply obtain $c$, the solution of $\frac{5}{x}=6x$ and $d$, the solution of $\frac{3}{x}=3x$ and integrate the function $5/x$ from a to c minus the integral of function $6x$ over the same range. Do the same for the other two terms ($3/x$ and $3x$) over range $b$ to $d$. (${a,b,c,d}\in R^+$)

$a=\frac{\sqrt{2}}{2},b=\sqrt{\frac{5}{3}},c=\sqrt{\frac{5}{6}},d=1$

Now, $S = area - (\int_{a}^{c}\frac{5}{x} - 6x dx ) - (\int_{b}^{d}\frac{3}{x} - 3x dx ) = [ln(2x)]_{a}^{b} - [ln(5x)-3x^2]_{a}^{c}-[ln(3x)-1.5x^2]_{b}^{d}$

Since $\ln(a)+\ln(b)=\ln(ab), \ln(a)-\ln(b)=\ln(a/b)$ then the above can be written as $\ln(\frac{b}{cd})+3c^{2}-3a^{2}+1.5d^{2}-1.5b^{2}=\ln(\sqrt{2})$

Since it is a hyperbola (mirror image on 4th quadrant), multiply the area between the two functions $5/x$ and $3/x$ by two. So, $2\ln(\sqrt{2})=\ln2$.

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  • $\begingroup$ i dont know why $\r$ doesnt work. $\endgroup$ – McPig Dec 6 '19 at 0:08

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