Integral over geometric region

Question

I'd like to calculate this integral

$$ \int_E y\ dydz $$ where $E = \{ (x,y,z) \in R^3 : z^2+6 < y^2 < 5z \}$

By hand i've got $\frac{1}{12}$ but i'm not sure, and i'd like to verify this integral and others.

I tried using ParametricRegion

ParametricRegion[{{y, z}, 6 + z^2 < y^2 < 5 z}]
Integrate[y, {z, y} \[Element] Region]

The result is very strange

(52 Sqrt[2/5])/3 - 14 Sqrt[3/5]

possible duplicate of How to plot and find the volume of a solid?, also Undocumented use of Integrate: Integrating over regions — Artes, Jul 09 '14 at 17:50
Artes, have you seen the specific geometric region? I don't know integration extremes! — apt45, Jul 09 '14 at 18:01
Your hand result is correct for y>0 . Note you have a mirrored region ( y<0 ) which cancels so the result should be zero. — george2079, Jul 09 '14 at 18:22
@george2079 thank you. It's right to consider just the right (or the left) side because i've to study a rotation solid and this is the plane shape that i've to rotate! thank you! Have you solved it by hand or with mathematica? — apt45, Jul 09 '14 at 18:25
uh.. you've reversed {y,z} in the region and integral.. (and I cut paste the error ) Fix that and it works (add y>0 to the region to get 1/12 ) — george2079, Jul 09 '14 at 18:31
@george2079 i tried with another function and domain: this is the input and the output http://i60.tinypic.com/eb8s5u.png Did i wrong something? — apt45, Jul 10 '14 at 08:39

score 2 · Accepted Answer · answered Jul 09 '14 at 18:43

The way you wrote it it obviously vanishes, since all the conditions are on $y^2$ while the integrand is $y$ (so whatever the $y>0$ bit contributes is cancelled by the $y<0$ bit).

If you meant for $y>0$ to be true, then eg

Integrate[y*Boole[z^2 + 6 < y^2] 
    Boole[y^2 < 5 z] Boole[0 < y], {y, -\[Infinity], \[Infinity]}, {z, -\[Infinity], \[Infinity]}]

give $1/12$.

But this sort of thing breaks down the moment you increase complexity.

Integral over geometric region

1 Answers1