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...
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?
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
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?
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?
The problem requires you to change the order of integration. The limits that are given have you integrating with respect to y first.