# Find the mass of the cylinder

• January 31st 2011, 02:50 PM
wopashui
Find the mass of the cylinder
Find the mass of the cylinder $S:0<=z<=h$, $x^2+y^2<=a^2$ if the density at the point $(x,y,z)$ is $\delta = 10z^4+6(x^2-y^2)^2$.

I think we should use the cylincal polar method to do this, however i have problem converting this into the polar form, which is $x=rcos\theta$, $y =rsin\theta$, $z=z$
• January 31st 2011, 03:36 PM
Ackbeet
You're having trouble converting the integrand into cylindrical coordinates? What happens when you plug in for x and y?
• January 31st 2011, 03:52 PM
HallsofIvy
You might find this trig identity helpful:

$cos^2(x)- sin^2(x)= sin(2x)$.

Perhaps you will remember the more basic identity
cos(a+ b)= cos(a)cos(b)- sin(a)sin(b).

Let a= b= x.
• January 31st 2011, 06:48 PM
wopashui
Quote:

Originally Posted by HallsofIvy
You might find this trig identity helpful:

$cos^2(x)- sin^2(x)= sin(2x)$.

Perhaps you will remember the more basic identity
cos(a+ b)= cos(a)cos(b)- sin(a)sin(b).

Let a= b= x.

so, after converting, I got $\int _\,^\!$ $\int _0\,^h\!$ $\int _0\,^{a^2}\!$ $10z^4+6rsin^2$(2 $\theta$ ) $dr dz d\theta$

is this correct? I'm still not sure with the integrand of $\theta$ however.
• February 1st 2011, 01:07 AM
Ackbeet
No, I don't think you've quite got it yet. Just plugging in gives:

$\delta=10z^4+6(x^2-y^2)^2=10z^4+6((r\cos(\theta))^2-(r\sin(\theta))^2)^2=10z^{4}+6(r^{2}\cos^{2}(\thet a)-r^{2}\sin^{2}(\theta))^{2}$

$=10z^{4}+6(r^{2}(\cos^{2}(\theta)-\sin^{2}(\theta)))^{2}=
10z^{4}+6(r^{2}\sin(2\theta))^{2}=10z^{4}+6r^{4}\s in^{2}(2\theta).$

You see how I work here: step-by-step. I don't often skip steps.

The other thing you have to worry about is that the volume differential in cylindrical components is $r\,dr\,d\theta\,dz,$ not just $dr\,d\theta\,dz.$

So, putting this all together, what does your integral look like now?
• February 1st 2011, 08:32 AM
wopashui
Quote:

Originally Posted by Ackbeet
No, I don't think you've quite got it yet. Just plugging in gives:

$\delta=10z^4+6(x^2-y^2)^2=10z^4+6((r\cos(\theta))^2-(r\sin(\theta))^2)^2=10z^{4}+6(r^{2}\cos^{2}(\thet a)-r^{2}\sin^{2}(\theta))^{2}$

$=10z^{4}+6(r^{2}(\cos^{2}(\theta)-\sin^{2}(\theta)))^{2}=
10z^{4}+6(r^{2}\sin(2\theta))^{2}=10z^{4}+6r^{4}\s in^{2}(2\theta).$

You see how I work here: step-by-step. I don't often skip steps.

The other thing you have to worry about is that the volume differential in cylindrical components is $r\,dr\,d\theta\,dz,$ not just $dr\,d\theta\,dz.$

So, putting this all together, what does your integral look like now?

yes, r is the jocobian, so what are the integrands here, i got r from 0 to $a^2$, z from 0 to h, how about $\theta$. is it 0 to 2pi?
• February 1st 2011, 08:35 AM
Ackbeet
Yes. The condition $x^{2}+y^{2}\le a^{2},$ with no explicit limits on the angle, implies that $0\le\theta\le 2\pi.$
• February 2nd 2011, 10:34 AM
wopashui
Quote:

Originally Posted by Ackbeet
Yes. The condition $x^{2}+y^{2}\le a^{2},$ with no explicit limits on the angle, implies that $0\le\theta\le 2\pi.$

i think i made a mistake finding the integrand of r, it should be in terms of $\theta$, not $0 to a^2$
• February 2nd 2011, 10:47 AM
Ackbeet
Show your work, and I'll take a look.