Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
Have you tried drawing a picture? –  Harald Hanche-Olsen Apr 18 at 22:45
    
For the record, yes, I did draw it out. –  user1833028 Apr 19 at 5:11

1 Answer 1

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.