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Double Integration

Ok so I am stuck on the question highlighted.
http://img704.imageshack.us/img704/681/calc0.png

I have it integrating over the rectangular domain : 0<=x<=4 and 0<=y<=2 so I have got

∫04(∫02x1+y5dy)dx \int^4_0 (\int^2_0 \dfrac{x}{1+y^5} dy) dx


But I'm stuck on how to integrate this.
Original post by Jooeee
Ok so I am stuck on the question highlighted.
http://img704.imageshack.us/img704/681/calc0.png

I have it integrating over the rectangular domain : 0<=x<=4 and 0<=y<=2 so I have got

∫04(∫02x1+y5dy)dx \int^4_0 (\int^2_0 \dfrac{x}{1+y^5} dy) dx


But I'm stuck on how to integrate this.


This isn't a rectangular domain, the question says: R is region bounded by y-axis, y=sqrt(x) and y=2.

If you sketch this region out, maybe that will give you a better idea
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Original post by ForGreatJustice
This isn't a rectangular domain, the question says: R is region bounded by y-axis, y=sqrt(x) and y=2.

If you sketch this region out, maybe that will give you a better idea


Ok yeah i sketched it out, and saw it wasn't rectangular - was being a retard lol. I still think I have got the right limits though, so how do I proceed. Can I just directly integrate, or because its non-rectangular do I need to use some other method. Because I'm pretty sure x/1+y^5 integrated is going to give a really messy answer :tongue:
You can just integrate directly, yes. With the correct limits, as the questions hints at, one order of integration is significantly easier than the other!
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Original post by ForGreatJustice
You can just integrate directly, yes. With the correct limits, as the questions hints at, one order of integration is significantly easier than the other!


Are my limits correct?

And integrating first w.r.t x is the much easier method yea?
The limits you gave in the OP are wrong... since those limits would have you integrate over the whole rectangle, which we don't want! However yes, the x integral should be easy
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Original post by ForGreatJustice
The limits you gave in the OP are wrong... since those limits would have you integrate over the whole rectangle, which we don't want! However yes, the x integral should be easy


Ok, I think my x-limits are correct. Can I give the y limits as being 0 and x \sqrt x ?

Edit: I used

∫02(∫0y2x1+y5dx)dy \int^2_0 (\int^{y^2}_0 \dfrac{x}{1+y^5} dx) dy

I got an answer out of it, not sure if its right though.
(edited 13 years ago)
Those y limits are wrong... but close. Look at the picture of the region again, only at x=0 is y=0 included, so what we want in the region Above the curve, not below

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