How to find the centroid of this tricky surface?

Question

Problem: The solid bounded by the surface $z=\sin(x), x=0,x=\pi,y=0,z=0,$ and $y+z=1,$ where the density $\rho=1.$

My Attempt: I know that in order to find the centroid one first must compute the mass of the solid, which is given by the following expression $$m=\int\int\int \rho dxdydz.$$ However, I am unable to come up with the bounds for this triple integral. I have come up with the bounds for the variables $x$ and $y$ but for $z$ I am still confused.
$$\int_{0}^{\pi}\int_{0}^{1}\int_{?}^{?} \rho dxdydz$$

Any hints/suggestions would be much appreciated.

Answer 1

If $\rho(x,y,z)=1$ then this computation simplifies to be a double integral.

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The body is bounded by $z=0$ from below and by the following function of $x$ and $y$ from above:

$$z(x,y)=\begin{cases} \sin(x)&\text{ if }& 0\le y\le 1-\sin(x)&\text{ and }& 0\le x\le \pi\\ 1-y&\text{ if }& 1-\sin(x)<y\le 1&\text{ and }& 0\le x\le \pi\\ 0&\text{ otherwise} \end{cases}$$

as it is shown in the following figure

The volume of this body (its mass if $\rho=1$) is given by the double integral

$$\iint_{[0,\pi]\times[0,1]} z(x,y)\ dx\ dy=$$ $$=\int_0^{\pi}\left[\int_0^{1-\sin(x)}\sin(x)\ dy +\int_{1-\sin(x)}^11-y\ dy\ \right]dx=$$ $$=\int_0^{\pi}\left[\sin(x)\int_0^{1-\sin(x)}\ dy +\int_{1-\sin(x)}^1\ dy-\int_{1-\sin(x)}^1y\ dy\ \right] dx=\cdots$$

Answer 2

We have $0\le z\le \sin x$, $0\le x\le \pi$ and $0\le y\le1-z$. The $y$ bounds depend on $z$. The $x$ bounds depend on $x$. So put $x$ on the outside, $z$ inside and $y$ on the very inside to get $$\int_0^\pi\int_0^{\sin x}\int_0^{1-z}\,dy\,dz\,dx.$$