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$\int \int_{D} x dxdy $, where $D$ is a region of the plane.

I know how to calculate directly using cartesian coordinates, but in this exercise I have to do a change of variables.

Consider $D$ the region limited by the lines $y=-x+4,y=-x+2,y=2x,y=x$.

The change of variables that I choosed is probably not the best one (because of the integrals that I got later): polar coordinates. $x=r\cos(\theta)$,$y=r\sin(\theta)$.

Since it is limited by $x$ and $2x$, $\theta$ will vary from $\frac{\pi}{4}$ to $\arctan(2)$.

Since $r\sin(\theta) = -r\cos(\theta)+2 \Rightarrow r = \frac{2}{\sin(\theta)+\cos(\theta)}$ and $r\sin(\theta) = -r\cos(\theta)+4 \Rightarrow r = \frac{4}{\sin(\theta)+\cos(\theta)}$ , the integral is:

$\int_{\frac{\pi}{4}}^{\arctan2} \int_{\frac{2}{\sin(\theta)+\cos(\theta)}}^{\frac{4}{\sin(\theta)+\cos(\theta)}} r^{2}\cos(\theta)drd\theta = \int_{\frac{\pi}{4}}^{\arctan2} \frac{\cos(\theta)}{3} (\frac{64}{(\sin(\theta)+\cos(\theta))^{3}} - \frac{8}{(\sin(\theta)+\cos(\theta))^{3}} )d\theta = \frac{56}{3} \int_{\frac{\pi}{4}}^{\arctan2} \frac{\cos(\theta)}{\sin(\theta)+\cos(\theta)}d\theta$

which won't be equal to the area that I calculated via cartesian coordinates: $1$.

Where is my mistake? Could you suggest a more convenient coordinate change? Thanks.

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    $\begingroup$ $$\int dr \, r^2 = \frac13 r^3 \cdots$$ $\endgroup$ – Ron Gordon Jun 3 '16 at 0:06
  • $\begingroup$ I don't understand... where did I integrate $r^2$ wrong? (I added more steps to my post to clarify what I did) $\endgroup$ – user286485 Jun 3 '16 at 0:32
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I suggest $u=x+y$, $v=\frac yx$ since that makes the region a rectangle with sides parallel to the coordinate axes. Inverting this gives $$x={u\over v+1}\\y={uv\over v+1}$$ with $\left|{\partial(x,y)\over\partial(u,v)}\right|=\frac{u}{(v+1)^{2}}$.

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  • $\begingroup$ Thanks! I tried solving: $\int_{1}^{2} \int_{2}^{4} \frac{u}{v+1} \frac{u}{(v+1)^{2}} du dv = \int_{1}^{2} \int_{2}^{4} \frac{u^2}{(v+1)^3} dudv = \int_{1}^{2} \frac{1}{(v+1)^{3})} \frac{64-8}{3} dv = \frac{56}{3} \int_{1}^{2} (v+1)^{-3} dv = \frac{56}{3} (\frac{(v+1)^{-2}}{-2})_{1}^{2} = \frac{35}{27}$ but I didn't get $1$, as I did in cartesian coordinates... where is my mistake? $\endgroup$ – user286485 Jun 3 '16 at 1:53
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    $\begingroup$ @Alnitak Thanks for fixing my Jacobian. I get ${35\over27}$ both ways. Divide up the region into three parts: $\int_{\frac23}^1\int_{2-x}^{2x}x\,dy\,dx + \int_1^{\frac43}\int_x^{2x}x\,dy\,dx + \int_{\frac43}^2\int_x^{4-x}x\,dy\,dx$. $\endgroup$ – amd Jun 3 '16 at 2:25
  • $\begingroup$ Is this way of cartesian integrating right? $\int_{\frac{2}{3}}^{1} 2x - (2-x) dx + \int_{1}^{\frac{4}{3}} 2x - x dx + \int_{\frac{4}{3}}^{2} (4-x) - x dx $ Just saw my problem of forgetting $x$. Thanks $\endgroup$ – user286485 Jun 3 '16 at 4:29
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$$ dx\, dy = r \,dr d\theta, x=r \cos \theta, $$

Integrand is

$$ r ^ 2 \,\cos \theta \,dr d\theta. $$

Tackle $r$ part first, inside integral sign $ \rightarrow r^ 3/3.$

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  • $\begingroup$ I don't see where I integrated $r^2$ wrong... (Added more steps to post to clarify what I did) $\endgroup$ – user286485 Jun 3 '16 at 0:34

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