Three-dimensional Lebesgue-measure How can I compute $\int_B gd\lambda^3$ where $$g(x,y,z)=xyz$$ and $$B=\{(x,y,z)\in\mathbb R^3\vert x,y,z ≥ 0, x^2+y^2+z^2 ≤ R^2\},$$
$R>0$ arbitrary?
I have no clue on how to find the upper and lower limits of the integrals, can someone give me a hint please?
 A: I will calculate the integral component-wise. First we need to fix an order between the variables. As $g$ is symmetric, I will integrate $g$ w.r.t $x,y,z$ respectively. This means, for every $y,z$ I need to calculate the integral 
$$\int_{x^2+y^2+z^2\le R^2}x\,dx$$
as a function of $y,z$. As $x\ge0$, and $x^2$ is an increasing funtion in the positive part of the real line,
$$x^2+y^2+z^2\le R^2\iff x\in\left[0,\sqrt{R^2-y^2-z^2}\right]$$
Similarly, for fixed $z$, there exists $x\in \mathbb{R}$ such that 
$$x^2+y^2+z^2\le R^2\iff y\in\left[0,\sqrt{R^2-z^2}\right]$$
Note that, as we first integrate against $x$, we are only fixing $z$, when integrating against $z$. Lastly, when choosing $z$, we have no constraint, except the obvious one $z\le R$.
For a more general case, let $A$ be Lebesgue measurable set. For all $y,z\in\mathbb{R}$, define 
$$A_{y,z}=\{x\in\mathbb{R}:(x,y,z)\in A\}$$
Then, by Fubini's Theorem:
$$\int_A xyz\,d\lambda=\int_{\mathbb R^3}xyz1_{A}\,d\lambda=\int_{-\infty}^\infty z \int_{-\infty}^\infty y \int_{-\infty}^\infty x1_A\,dx\,dy\,dz=\int_{-\infty}^\infty z \int_{-\infty}^\infty y \int_{A_{y,z}} x\,dx\,dy\,dz$$
, where $1_A$ denotes the characteristic function of the measurable set $A$.
Finding $A_{y,z}$ is obvious for our case, which gives the bound on $x$. As 
$$\int_{-\infty}^\infty y \int_{A_{y,z}} x\,dx\,dy$$
lies inside the integral against $z$, we are taking these integrals for a fixed $z$. So, we want the answer as a function of $z$. Moreover, for fixed $z$, there is no $(x,y,z)\in B$ such that $y>\sqrt{R^2-z^2}$. So, 
$$\int_{-\infty}^\infty y \int_{A_{y,z}} x\,dx\,dy=\int_0^\sqrt{R^2-z^2} y \int_{A_{y,z}} x\,dx\,dy$$
$$\int_B xyz\,dx\,dy\,dz=\int_0^Rz\int_0^{\sqrt{R^2-z^2}}y\int_0^{\sqrt{R^2-y^2-z^2}}x\,dx\,dy\,dz$$
$$=\frac12\int_0^Rz\int_0^{\sqrt{R^2-z^2}}y(R^2-z^2-y^2)\,dy\,dz$$
$$=\frac12\int_0^Rz\left((R^2-z^2)\int_0^{\sqrt{R^2-z^2}}y-\int_0^{\sqrt{R^2-z^2}}y^3\right)\,dy\,dz$$
$$=\frac12\int_0^Rz\left(\frac12(R^2-z^2)^2-\frac14(R^2-z^2)^2\,dz\right)$$
$$=\frac18\int_0^Rz(R^2-z^2)^2\,dz=\frac18\int_0^RR^4z-2R^2z^3+z^5\,dz$$
$$=\frac18\left(R^4\cdot\frac12R^2-2R^2\cdot\frac14R^4+\frac16R^6\right)=\frac18R^6(\frac12-\frac12+\frac16)=\frac1{48}R^6$$
