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 one year ago
Hi all! Need serious Multivariable Calc help.
Right now we're learning practical applications of double integrals:
An annulus with inner radius r=1 and outer radius r=2 has density equal to delta(x,y)=3*(y+sqrt(x^2+y^2))/pi. Compute the mass of the annulus using polar coordinates.
I get an answer, but I'm almost sure it's wrong.
 one year ago
Hi all! Need serious Multivariable Calc help. Right now we're learning practical applications of double integrals: An annulus with inner radius r=1 and outer radius r=2 has density equal to delta(x,y)=3*(y+sqrt(x^2+y^2))/pi. Compute the mass of the annulus using polar coordinates. I get an answer, but I'm almost sure it's wrong.

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oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1to find the area we must integrate \(\delta\) over the annulus, which is clearly represented in polar coordinates using \(\{(r,\theta):1\le r\le 2\}\)

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1so notice we must rewrite \(\delta\) in terms of \(r,\theta\); this is no problem as \(r=\sqrt{x^2+y^2}\) and \(y=r\cos\theta\):$$\delta(r,\theta)=\frac3\pi r\ (1+\cos\theta)$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1now we set up our integral:$$m=\int_0^{2\pi}\int_1^2\frac3\pi r\ (1+\cos\theta)\ dr\ d\theta=\frac3\pi\left(\int_0^{2\pi}(1+\cos\theta)\ d\theta\right)\left(\int_1^2r\ dr\right)$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1so clearly \(\displaystyle\int_1^2r\ dr=\frac12(2^21^2)=\frac32\) and the other integral turns out to be rather simple as well: $$\int_0^{2\pi}(1+\cos\theta)\ d\theta=\bigg[\theta+\sin\theta\bigg]_0^{2\pi}=2\pi+\sin2\pi\sin0=2\pi$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1thus \(\displaystyle m=\frac3\pi\cdot\frac32\cdot2\pi=9\)

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1Oh wow Thank you so much!!!

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1notice it's safe to rewrite the integral as I did above:$$\int\int f(x)g(y)\ dx\ dy=\int g(y)\int f(x)\ dx\ dy=\int f(x)\ dx\cdot\int g(y)\ dy$$

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1yeah we saw that in class about 2 days ago.

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1only because there were no factors that depended on both \(r,\theta\)... the \(1+\cos\theta\) only depends on \(\theta\) so we may pull it out as a constant from the \(dr\) integral; similarly, \(r\) depends not on \(\theta\) and so we may pull it out of the \(d\theta\) integral

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1ok, just wanted to clarify... I do it all the time and I know it confuses the people I tutor IRL

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1@oldrin.bataku wait...isn't y = r sin theta?

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1oops  good catch. I don't think it'll make a difference though

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1ok I'll double check the work and get back 2 ya

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1it won't make a difference since \(\cos(t)\) and \(\sin(t)\) have the same area over \([0,2\pi]\)... all this means is I picked a slightly different coordinate system:dw:1389663840845:dwthis is the regular polar coordinate system

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1here's my coordinate system:dw:1389663889927:dw

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1oh that makes sense. just another question: when converting to a polar integral, doesn't dxdy become rdrdtheta?

alffer1
 one year ago
Best ResponseYou've already chosen the best response.1so the final answer is 14.

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1oops, correcting for that we get $$\int_1^2 r^2\ dr=\frac13(2^31^3)=\frac73$$ergo our final result is \(\displaystyle\frac3\pi\cdot\frac73\cdot2\pi=14\)
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