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Calculate double integer

  • Thread starter kristink08
  • Start date
  • #1
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.

Homework Statement



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

Homework Equations





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...
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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?
 
  • #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
 
  • #4
Dick
Science Advisor
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618
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?
 
  • #5
I have no clue...:S
 
  • #6
Dick
Science Advisor
Homework Helper
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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?
 
  • #7
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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