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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.
 3 months ago
 3 months 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.
 3 months ago
 3 months ago

This Question is Closed

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

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

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

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

oldrin.batakuBest ResponseYou've already chosen the best response.1
thus \(\displaystyle m=\frac3\pi\cdot\frac32\cdot2\pi=9\)
 3 months ago

alffer1Best ResponseYou've already chosen the best response.1
Oh wow Thank you so much!!!
 3 months ago

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

alffer1Best ResponseYou've already chosen the best response.1
yeah we saw that in class about 2 days ago.
 3 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
only 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
 3 months ago

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

alffer1Best ResponseYou've already chosen the best response.1
@oldrin.bataku wait...isn't y = r sin theta?
 3 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
oops  good catch. I don't think it'll make a difference though
 3 months ago

alffer1Best ResponseYou've already chosen the best response.1
ok I'll double check the work and get back 2 ya
 3 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
it 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
 3 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
here's my coordinate system:dw:1389663889927:dw
 3 months ago

alffer1Best ResponseYou've already chosen the best response.1
oh that makes sense. just another question: when converting to a polar integral, doesn't dxdy become rdrdtheta?
 3 months ago

alffer1Best ResponseYou've already chosen the best response.1
so the final answer is 14.
 3 months ago

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