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1.

Find the area integral of $ \ f(x,y)=x^2 + y^2 \ $ in $ \ D=\{(x,y)\in\Bbb{R}|0\le x-2y\le2, |3x-y|\le1\} \ $ when we do a substitution $ \ u=x-2y \ $ and $ \ v=3x-y \ $

I did the calculations like this. Does it seem right?

$x=\frac 15(2v-u)$ and $y=\frac 15(v-3u)$

$dxdy=\frac{1}{\begin{vmatrix}1&-2\\3&-1\end{vmatrix}}dudv=\frac 15dudv$

$$\frac 15 \int_{-1}^{1}\int_0^2\frac{1}{25}(2v-u)^2+\frac{1}{25}(v-3u)^2dudv=\frac {12}{25} $$

2.

Also similar. $$\int_{-1}^0 \int_{-\sqrt{1-x^2}}^0 \frac{2}{1+\sqrt{x^2+y^2}}dxdy$$

I think I should go to polar coordinates and $x=-\sqrt{1-x^2}$ so $x=-\frac{1}{\sqrt2}$ and $x=0$. Then $\phi\in[5\pi/4, 3\pi/2]$ or because symmetry $\phi\in[0, \pi/4]$ $$\int_{0}^{\pi/4} \int_0^1 \frac{2r}{1+r}drd\phi$$

And now just integrate?

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  • $\begingroup$ In the second line, $v$ is $3x-y$, not $2x-y$. $\endgroup$
    – Etienne
    Commented Feb 23, 2014 at 20:52
  • $\begingroup$ Oh yeah thanks! Does the calculation and methods seem right? $\endgroup$
    – ELEC
    Commented Feb 23, 2014 at 20:55
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    $\begingroup$ The method is perfectly all right. I did not check the calculation, but I'm rather confident. $\endgroup$
    – Etienne
    Commented Feb 23, 2014 at 20:55
  • $\begingroup$ I got another similar with polar coordinates. does it seem right also at first glance? $\endgroup$
    – ELEC
    Commented Feb 24, 2014 at 11:08
  • $\begingroup$ This time, no. The domain of integration wrt $(r,\phi)$ is not correct. $\endgroup$
    – Etienne
    Commented Feb 24, 2014 at 12:54

1 Answer 1

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I confirm the result for the first integral (which turns out to require more writing than it first appears). Using the vertices of the parallelogram, we can find its area as $ \ | \ \langle \ \frac{2}{5} , \frac{6}{5} \ \rangle \ \times \ \langle \ \frac{4}{5} , \frac{2}{5} \ \rangle \ | \ = \ \frac{20}{25} \ = \ \frac{4}{5} \ $ . Since the integrand function is less than 1 over most of the domain, it is reasonable that we obtain a somewhat smaller value for the integral.

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For #2, judging from the limits of integration for your integral in Cartesian coordinates, the domain is the third-quadrant sector of the circle of radius 1 center on the origin. So, in polar coordinates, the integration runs over the intervals $ \ [0,1] \ $ for radius and $ \ [\pi, \frac{3 \pi}{2}] \ $ for angle. The area integral is then

$$\int_{\pi}^{3 \pi/2} \int_0^1 \ \frac{2r}{1+r} \ \ dr \ d\phi \ \ = \ \ \int_{\pi}^{3 \pi/2} d\phi \ \ \int_0^1 \ 2 \ - \ \frac{2}{1+r} \ \ dr \ $$

$$ \frac{\pi}{2} \ \cdot \ 2 \ \cdot \ ( \ r \ - \ \ln |1+r| \ ) \ \vert_0^1 \ = \ \pi \ ( \ 1 \ - \ \ln 2 \ ) \ \approx \ 0.31 \pi \ \ . $$

The area of the domain is $ \ \frac{\pi}{4} \ , $ and the value of the integrand function over the quarter-circle is only somewhat larger than 1 over most of the area, so this result for the integration is credible.

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