# Triple integral over a region bounded by surfaces.

The problem is to calculate $$I=\underset{S}{\iiint}x^2\;dV,$$ where $S$ is the region bounded by $x+y^2=1$, $x+z^2=1$ and the plane $zy$.

In my conclusions (that follows from this reasoning and probably are wrong), the value of the integral in the first octant is

$$J=\int_0^1\int_0^{\sqrt{1-x}}\int_0^{\sqrt{1-x}}x^2\;dz\;dy\;dx=\frac{1}{12}$$

and thus $I=2J=1/6$.

However, the book says that the answer is $1/3$. So, could someone tell me what is wrong and what is the correct way of write $I$ as an interated integral?

Thanks.

• Why are you only doubling it? Your region consists of four octants, not two. – Semiclassical Sep 4 '14 at 4:01
• @Semiclassical You're right. Thanks. – Pedro Sep 4 '14 at 4:11