Can you show me how to change the order of integration to the following integral (how to find the new limits of integration)?

$\int_{0}^{2}\int_{\sqrt{y}}^{1} (x^2+y^3x) dxdy$

The double integral is calculated over the closed region $D$ which can be represented as $$D=\big\{(x,y)\in\mathbb{R}\;|\; \sqrt{y}\leqslant {x}\leqslant 1, \; 0\leqslant {y}\leqslant 2\big\}\,.$$ Can you represent the same region $D$ in such way, such that the variable $x$ takes values from $0$ to $1$ and no from $\sqrt{y}$ to $1$?

(Try to draw $D$.)

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Grigorios Kostakos

I tried drawing D but I got confused. I represent $x=\sqrt{y}$ as $y=x^2$ and plot the lines $y=2$, $x=1$. It seems to me that D is divided in two sections: $0\leq x\leq1$ , $0\leq y\leq x^2$ and $1\leq x\leq\sqrt{2}$ , $x^2\leq y\leq 2$ , because the curve intersects the vertical line. Is this possible?

Attachment:

Andrew, I apologize. I wasn't careful and I consider $D$ to be the region $D_1$ in the scheme.
\[D=\big\{(x,y)\in\mathbb{R}\;|\; \sqrt{y}\leqslant {x}\leqslant 1, \; 0\leqslant {y}\leqslant 2\big\}\] is not a representation of $D_1\cup D_2$. $x$ must be $\leqslant 1$ and for $D_1\cup D_2$ this not the case.

So, must be a typo in the upper and lower limits of the integrals!

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Grigorios Kostakos

Oh, I see...The example was taken from Marsden-Tromba's Vector Calculus. These books aren't the gospel truth after all!

Marsden-Tromba's Vector Calculus is a good book (not "the gospel truth" for me) but even the masterpieces did not escape entirely from typos!

P.S. So, I suppose that your question about this double integral was answered (?)

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Grigorios Kostakos

Yes, this typo confused me. Thank you again for everything you've done!