Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

alffer1

  • 11 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.

  • This Question is Closed
  1. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  2. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 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. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    now we set up our integral:$$m=\int_0^{2\pi}\int_1^2\frac3\pi r\ (1+\cos\theta)\ dr\ d\theta=\frac3\pi\left(\int_0^{2\pi}(1+\cos\theta)\ d\theta\right)\left(\int_1^2r\ dr\right)$$

  4. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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: $$\int_0^{2\pi}(1+\cos\theta)\ d\theta=\bigg[\theta+\sin\theta\bigg]_0^{2\pi}=2\pi+\sin2\pi-\sin0=2\pi$$

  5. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thus \(\displaystyle m=\frac3\pi\cdot\frac32\cdot2\pi=9\)

  6. alffer1
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh wow Thank you so much!!!

  7. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    notice it's safe to rewrite the integral as I did above:$$\int\int f(x)g(y)\ dx\ dy=\int g(y)\int f(x)\ dx\ dy=\int f(x)\ dx\cdot\int g(y)\ dy$$

  8. alffer1
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah we saw that in class about 2 days ago.

  9. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 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

  10. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok, just wanted to clarify... I do it all the time and I know it confuses the people I tutor IRL

  11. alffer1
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @oldrin.bataku wait...isn't y = r sin theta?

  12. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oops -- good catch. I don't think it'll make a difference though

  13. alffer1
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok I'll double check the work and get back 2 ya

  14. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 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:dw|this is the regular polar coordinate system

  15. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    here's my coordinate system:|dw:1389663889927:dw|

  16. alffer1
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh that makes sense. just another question: when converting to a polar integral, doesn't dxdy become rdrdtheta?

  17. alffer1
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so the final answer is 14.

  18. oldrin.bataku
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oops, correcting for that we get $$\int_1^2 r^2\ dr=\frac13(2^3-1^3)=\frac73$$ergo our final result is \(\displaystyle\frac3\pi\cdot\frac73\cdot2\pi=14\)

  19. alffer1
    • 11 months ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sweet. Thanks.

  20. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.