Find the volume under surfaces. The first ones were fairly simple: h(x,y)=xy^2, 0 leq x leq 1, 1 leq y leq 2

kalkulusk2

kalkulusk2

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2022-08-21

Calculating limits of integrals of volume
I have a few problems to do with finding the volume under surfaces. The first ones were fairly simple:
h ( x , y ) = x y 2 , 0 x 1 , 1 y 2
V = 1 2 0 1 0 x y 2 d z d x d y = 1 2 0 1 x y 2 d x d y = 1 2 1 2 y 2 d y = 7 6
Step 2
However, on later questions the boundaries are given as:
h ( x , y ) = e x y , 0 x , y <
which to me implies: V = 0 0 exp ( x y ) d z d x d y .
However this evaluates to: V = [ e y ] . which is clearly divergent.
How should I calculate what the limits are for this volume?

Answer & Explanation

Harmony Horn

Harmony Horn

Beginner2022-08-22Added 11 answers

Step 1
The boundaries 0 x , y < mean that both x and y can have non-negative values:
0 x < 0 y < .
Step 2
So the lower bound in the outer integral (integrating w.r.t y) should be 0.

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