## reversing order of integration of double integral qns.

1. The problem statement, all variables and given/known data
pls refer to attached picture.

2. Relevant equations

3. The attempt at a solution

intially upper and lower limits are , x^2 < y< x^3 and -1<x<1
sketched y=x^2 and y= x^3. => sqrt(y) =x and cube root (y) = x
divide the area into 3 section.
new limits of dxdy
sqrt(y) <x< cube root (y) with 0<y<1 ,
and 0<x< sqrt(y) with 0<y< -1, ( for -ve x and +ve y portion of x^2 graph)
and cube root (y)< x< 0 with -1<y<0 ( for -ve x and -ve y portion of x^3 graph)

but the answer i have shows a different answer. guess i am wrong, but anyone can tell me which part?
attached is a graph i tried to draw( pardon my IT skills=P)
Attached Thumbnails

Attached Files
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 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 Recognitions: Homework Help Science Advisor Your picture is good, but you should emphasize that the region is also bounded by a vertical segment at x=(-1). Doesn't that mean some of your lower limits in the x integration need to be -1?
 hmm..to find limit of x integration we draw vertical line? so when i split the area to three sections, their upper limits is cube root y or sqrt y and the lower limit is 0? yea, but according to the answer i have, it uses limits of x integration as -1..and i dont understand why? could you kindly explain to me?

Recognitions:
Homework Help

## reversing order of integration of double integral qns.

Take your diagram and draw a segment connecting (-1,1) and (-1,-1). That's the boundary curve that determines your lower limit when you are doing the x<0 parts of the integration.

 opps..i was wrong, we should draw horizontal lines instead. and yup.. i get jus (for the -ve x portion) -1 < x < sqrt y with 0< y< 1 and -1 < x < -x^3 with -1< y < 0 using the these lines as a guide..
 opps i only can attach files when i post a thread? cant seem to be able to upload the edited graph here??

Recognitions:
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 Quote by blursotong opps..i was wrong, we should draw horizontal lines instead. and yup.. i get jus (for the -ve x portion) -1 < x < sqrt y with 0< y< 1 and -1 < x < -x^3 with -1< y < 0 using the these lines as a guide..
You'll need to express x limits in terms of y. And for both of them, the upper limit should be a negative number, like your graph shows.

 yup.i got them now. i get (jus for the -ve x portion) -1 < x < -sqrt y with 0< y< 1 and -1 < x < -cube root y with -1< y < 0 thanks alot for your guidance=D

Recognitions:
Homework Help