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How can I solve this double integral? $$\iint_{D}\frac{1}{x+6}\mathrm{d}x\mathrm{d}y$$ Where $D$ is the region between $y$-axis and the parametric curve $$ \begin{cases} x(t)=t-t^3\\ y(t)=2t-t^2 \end{cases} \qquad t\in [0,1]$$

I have never done this kind of problems, so I don't know how to start. Please help me.

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Using Fubini's theorem, you'll need "two lines" and "two curves". That is in this case, $y$ bounded between two values $0 \leq y \leq 1$.

Taking $y = 0$ and solving for $t$ you'll get $t=0 \, \lor \, t=2$. Since $0\leq t \leq1$ take $t=0$.

Taking $y = 1$ and solving for $t$ you'll get $t=1$

Therefore the bounds $0\leq y\leq1$ are equivalent to $0\leq t\leq1$.

The bounds of $x$ would be $$0\leq x\leq t-t^3$$

The integral can be now expressed as:

$$\int_{0}^{1}\int^{t-t^3}_0 \frac{1}{x+6}dx\,dt$$

Which can be computed one integral at a time.

$$\int^{t-t^3}_0 \frac{1}{x+6}dx=\ln(t-t^3+6)-\ln(6)$$

Taking the second integral

$$\int_{0}^{1}[\ln(t-t^3+6)-\ln(6)]dt = -3 + \ln{8} - 2\sqrt{2}\arctan(\frac{1}{\sqrt{2}})+2\sqrt{2}\arctan(\sqrt{2}) \approx 0.0406455$$

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  • $\begingroup$ Are you saying that $D$ is a normal region respect to $y-axis$? $\endgroup$
    – Aoeilda
    Feb 14, 2016 at 16:49
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    $\begingroup$ Fubini's theorem says you can change the order of integration. What that implies is that you can define the domain $D$ in different means. And naturally you would want to define $D$ in the way that allows you to integrate more easily. If you graph the curve, you'll quickly see what I mean. $\endgroup$
    – ex.nihil
    Feb 14, 2016 at 18:13
  • $\begingroup$ So I don't have to use the Green's Theorem? $\endgroup$
    – Aoeilda
    Feb 15, 2016 at 8:51
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    $\begingroup$ Well you could. You'd be converting it to a line integral which is perfectly fine. It's merely two different ways of looking at the problem. But usually Fubini's theorem is, by default, a way to get away from nasty calculaiton by redfining variables, changing the order of integration, etc. It's toying enough with the double integral expression to make it as easy as possible (for example you could also change coordinate systems if it is more convenient, not in this case though). $\endgroup$
    – ex.nihil
    Feb 15, 2016 at 11:12

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