Question

Find the mass of a cylinder \(0\leq r\leq 5,0\leq z\leq 10\), if its density at (x,y,z) is \(\sqrt{x^2+y^2}\)
Please, help

Answer 1

@IrishBoy123

Answer 2

for small volume element in cylindrical \(dV = r dr d \theta dz\)

mass of this is vol * density, so
\(dm = \rho dV\)

\(= \rho r dr d \theta dz\)

and \(rho = r\)

so \(= r^2 dr d \theta dz\)

do you agree?

Answer 3

ie integrate that

Answer 4

Yup

Answer 5

ok, so, \(0\leq \theta\leq 2\pi\)

\(0\leq r\leq 5\)

\(0\leq z\leq10\)
right?

Answer 6

yes

Answer 7

nope

Answer 8

the function should be r^(3/2) drdthetadz, right?

Answer 9

because rho at (x,y,z) = sqrt (x^2+y^2)= r
my bad :(

Answer 10

i'm gonna type it up

\(\int_{\theta = 0}^{ 2 \pi} \int_{z = 0}^{10} \int_{r = 0}^{5} r^2 dr dz d \theta\)

https://www.wolframalpha.com/input/?i=int_%7Btheta+%3D+0%7D%5E%7B+2+pi%7D+int_%7Bz+%3D+0%7D%5E%7B10%7D+int_%7Br+%3D+0%7D%5E%7B5%7D++r%5E2+dr++dz+d+theta

Answer 11

yes, it is. omg. you are so so so good. How can you remember all of them??

Answer 12

lol!!

Answer 13

i suspect you know this but indulge me!!! latexing again

Answer 14

\(\int_{\theta = 0}^{ 2 \pi} \int_{z = 0}^{10} \int_{r = 0}^{5} r^2 dr dz d \theta\)

\( = \int_{\theta = 0}^{ 2 \pi} d \theta * \int_{z = 0}^{10} dz * \int_{r = 0}^{5} r^2 dr \)

so you can do these in your head!!

how cool is that!!

Answer 15

yes, it is so simple now. :)
Thanks a ton, friend.