anonymous
  • anonymous
Use spherical coordinates to evaluate \int_Q (2+x^2+y^2+z^2)^(-1/2) dV where Q is the region above the xy-plane outside x^2+y^2+z^2=1 and inside x^2+y^2+z^2=16. Any help regarding setting up the limits would be appreciated.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
experimentX
  • experimentX
\[ \int_Q (2+x^2+y^2+z^2)^{-1/2} dV \] Change \( x^2 + y^2 + z^2 = r^2\) and \( dV = r^2 d\phi d\theta dr\) Rest should be easy, For reference: http://mathworld.wolfram.com/SphericalCoordinates.html
anonymous
  • anonymous
are you sure that's the right dV ?
anonymous
  • anonymous
oh\[dV=\rho^2\sin\phi d\rho d\phi d\theta\]anyway, for the bounds we have the region lying between two spheres\[x^2+y^2+z^2=\rho^2\]and between the two equations we have\[1\le\rho\le4\]does that make sense to you?

Looking for something else?

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

More answers

anonymous
  • anonymous
@CaptainSparkles still with me?
anonymous
  • anonymous
Ahh.. yes. Thanks, its just hard for me to get a hand of these but once I see someone else do it I understand.
anonymous
  • anonymous
good, so what are the bounds on phi and theta?
anonymous
  • anonymous
Sorry experiment, I wish I could give you best answer also.
anonymous
  • anonymous
I did for you :)
anonymous
  • anonymous
theta would be 0 to 2pi correct?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and phi is 0 to pi/2?
anonymous
  • anonymous
you got it :) ok I think you can take it from here, good job.
anonymous
  • anonymous
Thanks.
anonymous
  • anonymous
welcome

Looking for something else?

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