A region, D, is bounded by these four curves: $$y=x^3-4$$ $$y=x^3+4$$ $$x+y=6$$ $$x+y=0$$
I want to compute this integral by using a suitable change of variables: $$\iint_D(3x^3+3x^2y+x+y) \, dA$$
Just by looking at expressions that repeat themselves I'd guess this could be a feasible change of variables: $$u=x^3$$ $$v=x+y$$
But I have no idea where to go from here. Some help would be appreciated!
I've divided $D$ in four areas (look the picture below)
$D_1=\{(x,y)|0\leq x \leq 1,4\leq y\leq x^3+4\}\\ D_2=\{(x,y)|1\leq x \leq 2,4\leq y\leq x+6\}\\ D_3=\{(x,y)|(y-4)^(1/3)\leq x \leq (y+4)^{1/3},y_0\leq y\leq 4\}\\ D_4=\{(x,y)|-y\leq x \leq (y+4)^{1/3},y_1\leq y\leq y_0\}$
Call $f(x,y)=3 x^3 + 3 x^2 y + x + y$
and the integral can be split in these four parts
$$\int_0^1\int_4^{x^3+4}\,f(x,y) \,dy\,dx+ \int_1^2\int_4^{x+6}\,f(x,y) \,dy\,dx+\\ \int_{y_0}^4\int_{\sqrt[3]{y-4}}^{\sqrt[3]{y+4}}\, f(x,y)\,dx\,dy+ \int_{y_1}^{y_0}\int_{-y}^{\sqrt[3]{y+4}}\,f(x,y)\,dx\,dy$$
Where $y_0$ is the real solution of the equation $x^3-4=-x$
and $y_1$ is the real solution of $x^3+4=-x$
I have got respectively the following values
$$\iint_D(3x^3+3x^2y+x+y)\,dA=3.87+213.4+20.06+101.54\approx 338.87$$
I did my best
Hope this helps
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