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I want to compute with a triple integral the volume of the object that is calculated with $z = \sqrt{y}, z = 0 , x=0$ and $x+y=1$. I managed to draw the object. Then I tried to find the limits of the integrals.

I think $0 , $x^2 and $0.

But with these limits I end up with a negative result which is not very good when trying to find volume.

Any ideas cause I am really confused right now.

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    I guess I was just worried about the $\sqrt{y}$ when $y$ is negative. I think it must be implicit to take y>0. In which case I agree with Andre Nicolas's bounds.2012-05-09

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There is an unstated but implicit assumption that $y \ge 0$, because of the bound on $z$. This forces $x \le 1$. Thus our triple integral can be expressed as the iterated integral $\int_{x=0}^1\left(\int_{y=0}^{1-x}\left(\int_{z=0}^{\sqrt{y}} dz \right)dy\right)dx.$ The ensuing computation poses no special difficulties. The result is $\dfrac{4}{15}$.

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    I found the same result so it must be correct. Thanks.2012-05-10