1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $z = 3.5$. $\endgroup$ – mike239x Mar 9 '18 at 11:40
  • $\begingroup$ @mike239x note that it should be $3<z_G<4$ $\endgroup$ – gimusi Mar 9 '18 at 11:43
  • $\begingroup$ @gimusi yeah, of course. $\endgroup$ – mike239x Mar 9 '18 at 11:45
  • 1
    $\begingroup$ Moreover, the result for the volume(mass) appears to be wrong, as well as calculation itself. $\endgroup$ – mike239x Mar 9 '18 at 11:47
  • $\begingroup$ @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. $\endgroup$ – mike239x Mar 9 '18 at 11:51
1
$\begingroup$

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.

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

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 $$\int_0^{2\pi}\cos\theta\,d\theta=\int_0^{2\pi}\sin\theta\,d\theta=0$$

$\endgroup$
1
$\begingroup$

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 $r$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.