Question

Find the center of mass (gravity) of a solid of constant density that is bounded by the parabolic
cylinder x = y
2
and the planes x = z, z = 0, and x = 1. (

Answer 1

The solid V and its projection on the $xy-$plane are given in Figures 01 and 02.

The lower and upper surfaces of the $V$ are planes $z=0$ and $z=x.$ Describe $V$ as a region

$$V=[(x,y,z):-1\leq y\leq 1, y^2\leq x\leq 1,0\leq z\leq x]$$

If the density function $f(x, y, z)=k, k=const,$ then the mass is

$$m=\iiint_VfdV=\displaystyle\int_{-1}^1\displaystyle\int_{y^2}^1\displaystyle\int_{0}^xkdzdxdy=$$

$$=k\displaystyle\int_{-1}^1\displaystyle\int_{y^2}^1(x-0)dxdy=k\displaystyle\int_{-1}^1\bigg[{x^2\over 2}\bigg]\begin{matrix} 1 \\ y^2 \end{matrix}\ dy=$$

$$={k\over 2}\displaystyle\int_{-1}^1(1-y^4)dy={k\over 2}\bigg[y-{y^5\over5}\bigg]\begin{matrix} 1 \\ -1 \end{matrix}=$$

$$={k\over 2}(1-{1\over 5}+1-{1\over 5})={4k\over 5}$$

Consider a function which is bounded by the curves $z=g_1(x, y)$  and $z=g_2(x, y)$

$$g_1(x, -y)=g_1(x, y), \ g_2(x, -y)=g_2(x, y)$$

Then we say that the region is symmetric about $y-$ axis and hence $M_{xz}=0,$ and the $y-$coordinate of the center of mass is zero $\bar{y}=0.$

$$M_{yz}=\iiint_VxfdV=\displaystyle\int_{-1}^1\displaystyle\int_{y^2}^1\displaystyle\int_{0}^xkxdzdxdy=$$

$$=k\displaystyle\int_{-1}^1\displaystyle\int_{y^2}^1x(x-0)dxdy=k\displaystyle\int_{-1}^1\bigg[{x^3\over 3}\bigg]\begin{matrix} 1 \\ y^2 \end{matrix}\ dy=$$

$$={k\over 3}\displaystyle\int_{-1}^1(1-y^6)dy={k\over 3}\bigg[y-{y^7\over7}\bigg]\begin{matrix} 1 \\ -1 \end{matrix}=$$

$$={k\over 3}(1-{1\over 7}+1-{1\over 7})={4k\over 7}$$

$$M_{xy}=\iiint_VzfdV=\displaystyle\int_{-1}^1\displaystyle\int_{y^2}^1\displaystyle\int_{0}^xkzdzdxdy=$$

$$={k\over 2}\displaystyle\int_{-1}^1\displaystyle\int_{y^2}^1x^2 dxdy={k\over 2}\displaystyle\int_{-1}^1\bigg[{x^3\over 3}\bigg]\begin{matrix} 1 \\ y^2 \end{matrix}\ dy=$$

$$={k\over 6}\displaystyle\int_{-1}^1(1-y^6)dy={k\over 6}\bigg[y-{y^7\over7}\bigg]\begin{matrix} 1 \\ -1 \end{matrix}=$$

$$={k\over 6}(1-{1\over 7}+1-{1\over 7})={2k\over 7}$$

Therefore, the centre of mass is

$$(\bar{x}, \bar{y},\bar{z})=\big({M_{yz}\over m},{M_{xz}\over m},{M_{xy}\over m}\big)=\big({5\over 7},\ 0,\ {5\over 14}\big)$$