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Surface Area Double Integral Problem?
I am having issues solving this problem, t s getting way too big on me so think I am making a mistake somewhere.
Here Is the problem as it is written:
Express the area of the given surface as an iterated double integral, and then find the surface area. The portion of the surface z=2x+y^2 that is above the triangular region with vertices (0,0), (0,1) and (1,1)
To set it up I ended up with int(sqrt(4+4y^2+1)dydx) My outer limits were from 0 to 1 while my inner limits were from x to 1
Any help would be greatly appreciated
1 Answer
- kbLv 78 years agoFavorite Answer
Try writing the bounds as x = 0 to x = y with y in [0, 1].
So, we have
∫(y = 0 to 1) ∫(x = 0 to y) √(4y^2 + 5) dx dy
= ∫(y = 0 to 1) y√(4y^2 + 5) dy
= (1/8)(2/3)(4y^2 + 5)^(3/2) {for y = 0 to 1}
= (1/12) [27 - 5^(3/2)].
I hope this helps!
