I did what you show me on the last post but my answer look weird. Can you check what I did wrong?
Let R be the region in the xy-plane which lies below the curve y = 𝑥^3, above the x-axis and for which 0 ≤ x ≤2. Compute the integral ∫∫R(𝑥+𝑦)𝑑A
a. integrating y first then x.
b. integrating x first then y.
part a.
[0,2]∫ [0,x]∫(x+y)dy dx=>[0,2]∫[xy+(y^2)/2][0,x]dx=>[0,2]∫x^2 +(x^2)/2dx(3/2)[0,2]∫x^2 dx=>(3/2)[(x^3)/3 [0,2]=>(3/2)*(8/3)=24/6=4
part b.
[0,8] ∫ [0,∛y]∫ x+y dx dy=>[0,8] ∫ [(x^2)/2 +yx][0,∛y]dy=>[0,8] ∫ (y^2/3)/2 +y^4/3 dy=>3(y^5/3)/10 +3(y^7/3)/7[0,8]=>96/10+384/7=2256/35
The two answers are the same. Should they have the same answer?
My bad. I made a typo in the first place. It should have been
[0,2]∫ [0,x^3]∫(x+y) dy dx = 544/35
On the second one, This gives the region above the curve, not below it. Draw the region in 3D and you can see that I should have written
[0,8] ∫ [∛y,2]∫ (x+y) dx dy = 544/35
So, sorry I misled you. Some analysis on your part may be necessary ...