Calculate double integer


by kristink08
Tags: double, integer
kristink08
kristink08 is offline
#1
May25-09, 03:55 PM
P: 3
This was a problem on a final test I took this april in Reykjavík University and I whould be greatful if you could help me with it.

1. The problem statement, all variables and given/known data

Let f(x,y)=2x*cos(y^4) be a function and let D be area in R^2 defined by 0≤x≤1 and x^(2/3)≤y≤1.
Calculate the double integer:
∫∫f(x,y)dA

2. Relevant equations



3. The attempt at a solution

∫dx∫2x*cos(y^4)dy
I have tried to use substitution but that doesn´t lead me anywhere.
I also tried to solve it this way...
∫dy∫2x*cos(y^4)dx which leads to...
∫dy*(x^2*cos(y^4)) and when I add in for x...
∫dy*cos(y^4) and if I use substitution now I will get...
1/4*∫cos(u)du and the final answer isn´t sufficient...
1/4*(sin(1)-sin(x^(8/3)))

I would be very greatful if you could help me...
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Dick
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#2
May25-09, 04:12 PM
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Carefully draw a sketch of the region. Now when you integrate dx, what will be the upper and lower limits of the integration in terms of y?
kristink08
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#3
May25-09, 06:00 PM
P: 3
the upper limits are 1 and lower limits are x^(2/3) in terms of y
and upper limits are 1 and lower limits are 1 in terms of x

Dick
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#4
May25-09, 06:52 PM
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Calculate double integer


That's not what my picture looks like. The integration dx goes along a horizontal line through the region. It's the part above (above being the positive y direction) the curve x^(2/3)=y. Want to try again?
kristink08
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#5
May26-09, 09:20 AM
P: 3
I have no clue...:S
Dick
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#6
May26-09, 09:30 AM
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The region inside of the square 0<=x<=1 and 0<=y<=1 above the curve y=x^(2/3). Pick a value of y and draw a horizontal line. Tell me what the x value is where it crosses the region. It looks to me like it will hit the y-axis first and then the curve, right?
Random Variable
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#7
May26-09, 09:45 AM
P: 116
The problem requires you to change the order of integration. The limits that are given have you integrating with respect to y first.


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