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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 $0<z<\sqrt{y}$ , $x^2<y<1-x$ and $0<x<\frac{\sqrt{5}}{2}-\frac{1}{2}$.

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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First $z$ goes $0$ to $\sqrt{y}$, then $y$ goes $0$ to $1-x$, then $x$ goes $0$ to $1$. – André Nicolas May 9 '12 at 23:12
A couple of things: (1) Do you also have a bound of $y = 0$? (2) I think your first limits $0<z<\sqrt{y}$ were ok, but the next two are funny. Why do you have $x^2<y$? – froggie May 9 '12 at 23:12
Nothing about $y=0$. I did $x^2<y$ cause I read in the book that you must take the projection to $z=0$ and find the limits as you do for a 2-integral. – srimas May 9 '12 at 23:17
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. – froggie May 9 '12 at 23:19
up vote 2 down vote accepted

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. – srimas May 10 '12 at 8:41

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