Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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.

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

to find the area we must integrate \(\delta\) over the annulus, which is clearly represented in polar coordinates using \(\{(r,\theta):1\le r\le 2\}\)
so notice we must rewrite \(\delta\) in terms of \(r,\theta\); this is no problem as \(r=\sqrt{x^2+y^2}\) and \(y=r\cos\theta\):$$\delta(r,\theta)=\frac3\pi r\ (1+\cos\theta)$$
now we set up our integral:$$m=\int_0^{2\pi}\int_1^2\frac3\pi r\ (1+\cos\theta)\ dr\ d\theta=\frac3\pi\left(\int_0^{2\pi}(1+\cos\theta)\ d\theta\right)\left(\int_1^2r\ dr\right)$$

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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: $$\int_0^{2\pi}(1+\cos\theta)\ d\theta=\bigg[\theta+\sin\theta\bigg]_0^{2\pi}=2\pi+\sin2\pi-\sin0=2\pi$$
thus \(\displaystyle m=\frac3\pi\cdot\frac32\cdot2\pi=9\)
Oh wow Thank you so much!!!
notice it's safe to rewrite the integral as I did above:$$\int\int f(x)g(y)\ dx\ dy=\int g(y)\int f(x)\ dx\ dy=\int f(x)\ dx\cdot\int g(y)\ dy$$
yeah we saw that in class about 2 days ago.
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
ok, just wanted to clarify... I do it all the time and I know it confuses the people I tutor IRL
@oldrin.bataku wait...isn't y = r sin theta?
oops -- good catch. I don't think it'll make a difference though
ok I'll double check the work and get back 2 ya
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
here's my coordinate system:|dw:1389663889927:dw|
oh that makes sense. just another question: when converting to a polar integral, doesn't dxdy become rdrdtheta?
so the final answer is 14.
oops, correcting for that we get $$\int_1^2 r^2\ dr=\frac13(2^3-1^3)=\frac73$$ergo our final result is \(\displaystyle\frac3\pi\cdot\frac73\cdot2\pi=14\)
sweet. Thanks.

Not the answer you are looking for?

Search for more explanations.

Ask your own question