Triple integral of a sphere bounded by a region The question is: Calculate the following integral
$$I = \iiint\limits_D  (x^2 + y^2 + z^2) \,dxdydz$$
where D is the region limited by $x+y+z\le2, x \ge 0, y \ge 0, z \ge 0$
My attempt was trying to picture the shape and I think it is a quater of a sphere that is bounded by the plane $ z = 2 -y-x$.
What I am not sure if if the resulting bounded shape is not just purely a quater of a sphere of ther is some additional region between the surface of the sphere and the plane.
Edit: I also tried using limits and realized I get an integral expression with powers of 3 which take long to calculate hence I thought there is a faster method possibly using spherical coordinates
Can you give me hints on how to attack the problem?
 A: \begin{aligned}\small\iiint_{D}{\left(x^{2}+y^{2}+z^{2}\right)\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z}&\small=\int_{0}^{2}{\int_{0}^{2-z}{\int_{0}^{2-y-z}{\left(x^{2}+y^{2}+z^{2}\right)\mathrm{d}x}\,\mathrm{d}y}\,\mathrm{d}z}\\ &\small=\int_{0}^{2}{\int_{0}^{2-z}{\int_{0}^{2-y-z}{x^{2}\,\mathrm{d}x}\,\mathrm{d}y}\,\mathrm{d}z}+\int_{0}^{2}{\int_{0}^{2-z}{y^{2}\left(2-y-z\right)\mathrm{d}y}\,\mathrm{d}z}+\int_{0}^{2}{z^{2}\int_{0}^{2-z}{\left(2-y-z\right)\mathrm{d}y}\,\mathrm{d}z}\\ &\small=\frac{1}{3}\int_{0}^{2}{\int_{0}^{2-z}{\left(2-y-z\right)^{3}\mathrm{d}y}\,\mathrm{d}z}+\frac{1}{12}\int_{0}^{2}{\left(2-z\right)^{4}\mathrm{d}z}+\frac{1}{2}\int_{0}^{2}{z^{2}\left(2-z\right)^{2}\,\mathrm{d}z}\\ &\small=\frac{1}{6}\int_{0}^{2}{\left(2-z\right)^{4}\,\mathrm{d}z}+\frac{1}{2}\int_{0}^{2}{z^{2}\left(2-z\right)^{2}\,\mathrm{d}z}\\ &\small=\frac{16}{15}+\frac{8}{15}\\ \small\iiint_{D}{\left(x^{2}+y^{2}+z^{2}\right)\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z}&\small=\frac{8}{5}\end{aligned}
A: I just saw this and hence may be late in responding but it is not necessary to compute the whole integral as is. The work can be simplified if we can make the symmetry argument.
Please note the region is $x, y, z \geq 0$ and $x + y + z \leq 2$. The region has symmetry wrt $x, y$ and $z$ and the integral of $x^2, y^2$ and $z^2$ would be same.
$\displaystyle I = \iiint_{D} (x^2 + y^2 + z^2) ~dx ~dy ~dz = 3 \iiint_{D} z^2 ~dx ~dy ~dz$
So the integral is,
$ \displaystyle I = 3 \int_0^2 \int_0^{2-z} \int_0^{2-y-z} z^2 ~dx ~ dy ~ dz = 3 \cdot \frac{8}{15} = \frac{8}{5}$
A: Compute$$\int_0^2\int_0^{2-x}\int_0^{2-x-y}x^2+y^2+z^2\,\mathrm dz\,\mathrm dy\,\mathrm dx.$$
