# Mass and center of mass of solid

Let $S$ be the part of the solid $z = 4 - x^2 - y^2$ where $z$ $\geq 3$, and let $T$ be the solid enclosed by $S$ and the plane $z = 3$. Assume $T$ has constant density equal to $1$. Calculate the mass and the coordinates of the center of mass of $T$.

So to find the mass I solved $$\int_0^{2\pi} \int_0^1 \int_3^4rdzdrd\theta = \pi$$ and then found the $z$ coordinate for the center of mass to be $$\frac1\pi\int_0^{2\pi} \int_0^1 \int_3^4rzdzdrd\theta = \frac72$$ Am I correct so far? Would it also be correct to asume that the senter of mass must lie on the $z$-axis so that the center of mass has the coordinates $(0,0,\frac72)$?

By the way, is it possible to solve this using only a double integral, or must a triple integral be used?

• I am pretty sure you made a mistake while finding center of mass -- just from imagining that solid it is clear that its center of mass lies below $z = 3.5$. – mike239x Mar 9 '18 at 11:40
• @mike239x note that it should be $3<z_G<4$ – gimusi Mar 9 '18 at 11:43
• @gimusi yeah, of course. – mike239x Mar 9 '18 at 11:45
• Moreover, the result for the volume(mass) appears to be wrong, as well as calculation itself. – mike239x Mar 9 '18 at 11:47
• @Pame Can you Include info on how you parameterize $T$? It's a bit hard to pinpoint the exact position of an error without this info. – mike239x Mar 9 '18 at 11:51

Yes it is the correct way to solve since by symmetry the center of mass lies on $z$ axes (I didn't check the detail of calculation).

Yes of course we can also use double or single integral by disk or shell method.

• How would you set up the double and single integrals? – Pame Mar 9 '18 at 11:42
• @Pame You can consider the revolution $of z=4-x^2$ bounded by $z=3$ around the $z$ axis by disk method. – gimusi Mar 9 '18 at 11:44

The triple integral is the easiest I think. You have the condition $$4-r^2=z\ge3$$ So that $r\le1$ is your upper bound for $r$. The lower bound is clearly $0$. Now for $z$, you were given that $3\le z\le4-r^2$, so the mass integral becomes $$m=\int_0^{2\pi}\int_0^1\int_3^{4-r^2}dz\,r\,dr\,d\theta=2\pi\int_0^1(1-r^2)r\,dr=\left.2\pi\left(-\frac14\right)(1-r^2)^2\right|_0^1=\frac{\pi}2$$ Then $$m\bar z=\int_0^{2\pi}\int_0^1\int_3^{4-r^2}z\,dz\,r\,dr\,d\theta=2\pi\int_0^1\frac12\left[(4-r^2)^2-9\right]r\,dr=\pi\left[\left(-\frac16\right)(4-r^2)^3-\frac92r^2\right]_0^1=\frac{5\pi}3$$ So $$\bar z=\frac{m\bar z}m=\frac{10}3$$ And this is between $3$ and $4$ as it should be. Your mistake lay in not having the correct upper bound for $z$. $\bar x=\bar y=0$ because when you do the $\theta$ integrals ou will find that $$\int_0^{2\pi}\cos\theta\,d\theta=\int_0^{2\pi}\sin\theta\,d\theta=0$$

First, the volume:

For fixed $z \in [3,4]$ the slice of $T$ you get by fixing $z$ is a circle $x^2+y^2 \leq 4-z$.

Its area is $\pi r^2$, where $r^2 = 4-z$.

Now the volume calculation goes as following: $$\int_3^4\pi\cdot(4-z)\ dz = \pi \cdot \left(4z-\frac{z^2}{2}\right)\bigg|_{z=3}^{z=4} = \pi\cdot((16-8)-(12-4.5)) = \frac{\pi}{2}$$

Second, the center of mass:

For each slice its center of mass is $(0,0,z)^T$ and the relative mass is $\pi\cdot(4-z)$.

To calculate the center of mass we do following:

$$\frac{2}{\pi}\cdot\int_3^4\pi(4-z)\cdot(0,0,z)^T\ dz = \frac{2}{\pi}\left(0,\ 0,\ \pi\cdot\int_3^4(4-z)z\ dz \right)^T =\ldots = \frac{2}{\pi}\left(0,\ 0,\ \pi\cdot\frac{5}{3} \right)^T = \left(0,\ 0,\ \frac{10}{3} \right)^T$$

Something like that. Of course you could also do the volume calculation with triple integral, like that: $$\int_{z =3}^4\int_{\theta =0}^{2\pi}\int_{r =0}^\sqrt{4-z} r\ dr d\theta d z = \ldots$$

And something similar with center of mass (once again, careful with integration limits for $r$).