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I have the following Double Integral:$\iint_Dx\cos y\space\mathrm{d}A$ where $a$ is bounded by $x=1,y=0,y=x^2$.

Interpreting this region as a Type one region, it is easy to conclude $R=\{(x,y)\mid 0\leq x \leq 1\space , \space 0 \leq y \leq x^2 \}$ is an appropriate bound for a double integral with respect to $\mathrm{d}y\,\mathrm{d}x$. However I cannot come up with a description that satisfies Describing the region as a Type II integral ( ie. $\mathrm{d}x\mathrm{d}y$).

My $y$-bound terms include attempts such as $0 \leq x \leq \sqrt{y}$.

However, double checking my computations on wolfram alpha, I cannot find bounds at all that indeed produce the answer.

That said, can ALL integrals be interpreted as type I or type II? If so, how can we prove an integral can or cannot be?

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Have you tried drawing a picture? – Harald Hanche-Olsen Apr 18 '14 at 22:45
For the record, yes, I did draw it out. – user1833028 Apr 19 '14 at 5:11

If by Type this or that integrals you mean changing the order of integration, you'd get

$$0\le x\le 1\;,\;\;0\le y\le x^2\;\;\longleftrightarrow\;\; 0\le y\le 1\;,\;\;\sqrt y\le x\le 1$$

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