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alffer1 Group Title

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.

  • 8 months ago
  • 8 months ago

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  1. oldrin.bataku Group Title
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    to find the area we must integrate \(\delta\) over the annulus, which is clearly represented in polar coordinates using \(\{(r,\theta):1\le r\le 2\}\)

    • 8 months ago
  2. oldrin.bataku Group Title
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    so notice we must rewrite \(\delta\) in terms of \(r,\theta\); this is no problem as \(r=\sqrt{x^2+y^2}\) and \(y=r\cos\theta\):$$\delta(r,\theta)=\frac3\pi r\ (1+\cos\theta)$$

    • 8 months ago
  3. oldrin.bataku Group Title
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    now we set up our integral:$$m=\int_0^{2\pi}\int_1^2\frac3\pi r\ (1+\cos\theta)\ dr\ d\theta=\frac3\pi\left(\int_0^{2\pi}(1+\cos\theta)\ d\theta\right)\left(\int_1^2r\ dr\right)$$

    • 8 months ago
  4. oldrin.bataku Group Title
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    so clearly \(\displaystyle\int_1^2r\ dr=\frac12(2^2-1^2)=\frac32\) and the other integral turns out to be rather simple as well: $$\int_0^{2\pi}(1+\cos\theta)\ d\theta=\bigg[\theta+\sin\theta\bigg]_0^{2\pi}=2\pi+\sin2\pi-\sin0=2\pi$$

    • 8 months ago
  5. oldrin.bataku Group Title
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    thus \(\displaystyle m=\frac3\pi\cdot\frac32\cdot2\pi=9\)

    • 8 months ago
  6. alffer1 Group Title
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    Oh wow Thank you so much!!!

    • 8 months ago
  7. oldrin.bataku Group Title
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    notice it's safe to rewrite the integral as I did above:$$\int\int f(x)g(y)\ dx\ dy=\int g(y)\int f(x)\ dx\ dy=\int f(x)\ dx\cdot\int g(y)\ dy$$

    • 8 months ago
  8. alffer1 Group Title
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    yeah we saw that in class about 2 days ago.

    • 8 months ago
  9. oldrin.bataku Group Title
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    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

    • 8 months ago
  10. oldrin.bataku Group Title
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    ok, just wanted to clarify... I do it all the time and I know it confuses the people I tutor IRL

    • 8 months ago
  11. alffer1 Group Title
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    @oldrin.bataku wait...isn't y = r sin theta?

    • 8 months ago
  12. oldrin.bataku Group Title
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    oops -- good catch. I don't think it'll make a difference though

    • 8 months ago
  13. alffer1 Group Title
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    ok I'll double check the work and get back 2 ya

    • 8 months ago
  14. oldrin.bataku Group Title
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    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:dw|this is the regular polar coordinate system

    • 8 months ago
  15. oldrin.bataku Group Title
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    here's my coordinate system:|dw:1389663889927:dw|

    • 8 months ago
  16. alffer1 Group Title
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    oh that makes sense. just another question: when converting to a polar integral, doesn't dxdy become rdrdtheta?

    • 8 months ago
  17. alffer1 Group Title
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    so the final answer is 14.

    • 8 months ago
  18. oldrin.bataku Group Title
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    oops, correcting for that we get $$\int_1^2 r^2\ dr=\frac13(2^3-1^3)=\frac73$$ergo our final result is \(\displaystyle\frac3\pi\cdot\frac73\cdot2\pi=14\)

    • 8 months ago
  19. alffer1 Group Title
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    sweet. Thanks.

    • 8 months ago
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