anonymous
  • anonymous
Find the volume of the region bounded by the sphere x^2+y^2+z^2=9 and the cone z=sqrt(x^2+y^2). I have already turned into polar coordinates with 0<=rho<=3, 0<=phi<=pi/4, and 0<=theta<=pi/2 integrating (rho^4)(sin(phi)) drho dphi dtheta. Can someone help me get the answer out of this? I keep coming up with the incorrect answer, so calculations are off somewhere (it is possible that i have my boundaries wrong...).
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • TuringTest
just off the bat, you that in spherical coordinates\[dV=rdrd\theta\phi\]you pick up an extra r
anonymous
  • anonymous
in spherical it picks up the p^2sinphi instead of the r right?, the r is only added in cylinderical coordinates.
anonymous
  • anonymous
p being rho

Looking for something else?

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

More answers

TuringTest
  • TuringTest
oh my bad, ok let me actually look at the question...
TuringTest
  • TuringTest
ok, why is theta only going to from 0 to pi/2 and not 2pi ?
TuringTest
  • TuringTest
it doesn't say "in the first octant"
anonymous
  • anonymous
no, I typed entire question and it is kicking my butt! I am so good at doing them individually, but putting them the sphere with the cone just throws me off.
TuringTest
  • TuringTest
I think it should be\[\int_{0}^{2\pi}\int_{0}^{\frac\pi2}\int_{0}^{3}\rho^2\sin\phi d\rho d\phi d\theta\]
TuringTest
  • TuringTest
I don't know why you cut everything in half is what I'm saying....
anonymous
  • anonymous
I began with 2pi but changed it when I found a similar problem on internet. I have found those to not be as helpful as I had hoped.
anonymous
  • anonymous
so 2pi would be the only incorrect thing you see in the integrand limits?
TuringTest
  • TuringTest
I also have going to pi/4 in for phi but that gives zero, which is the integral but not the actual vulume
TuringTest
  • TuringTest
phi to pi/2 rather
anonymous
  • anonymous
I kept getting final volume of 0 when I tried with pi/2. maybe my calculations are wrong
anonymous
  • anonymous
I have the choices between: A. (27/4)pi(2-sqrt2) B.9pi(2-sqrt3) C. 27/4)pi(2-sqrt3) D.9pi(2-sqrt2)
TuringTest
  • TuringTest
yes, because of the sin part I think that is normal because the object is symmetric about z, but that is clearly not the volume hey looky what I found pretty similar to this, but without a function being integrated over it and radius rho=1 http://tutorial.math.lamar.edu/Classes/CalcIII/TISphericalCoords.aspx
anonymous
  • anonymous
That is the page I found earlier that made me set the limits to pi/2 for theta and pi/4 for phi. Do you think I should stick with the integration limits that they have?
TuringTest
  • TuringTest
ok we know your rho bounds are right so let's narrow it down\[9\int_{0}^{?}\int_{0}^{?}\sin\phi d\phi d\theta\]that logic seems to hold in this situation to me, but since we want actual volume...
TuringTest
  • TuringTest
where did you get the bound pi/4 ?
anonymous
  • anonymous
i took rho(cosphi)=sqrt(9/2) (substituding in cosphi for z when you get to the part where you see z^2+z^2=rho
anonymous
  • anonymous
it ended up equaling pi/4
TuringTest
  • TuringTest
ok, so then my only suggestion is to change the bounds on theta to 2pi let's see if that gives an option...
anonymous
  • anonymous
I will try real quick and let you know. Thank you!
TuringTest
  • TuringTest
I get D that way
anonymous
  • anonymous
I have attached my work so far, but changing the theta integrand limits doesn't seem to get me to the right place. Could you take a look and see if I am doing something wrong along the way?
1 Attachment
TuringTest
  • TuringTest
I can't zoom in and see it :(
anonymous
  • anonymous
I tried re-doing the file, see if this helps...
1 Attachment
TuringTest
  • TuringTest
not much... I'm typing it out hold on
anonymous
  • anonymous
\[\int\limits_{0}^{2\pi}-243/5(\sqrt2)+243/5\] is where I ended up at.
TuringTest
  • TuringTest
\[\int_{0}^{3}\int_{0}^{\frac\pi4}\int_{0}^{2\pi}\rho^2\sin\phi d\rho d\theta d\phi=9\int_{0}^{\frac\pi4}\int_{0}^{2\pi}\sin\phi d\theta d\phi\]\[=18\pi\int_{0}^{\frac\pi4}\sin\phi d\phi =-18\pi\cos\phi|_{0}^{\frac\pi4}\]\[=-18\pi(\frac{\sqrt2}2-1)=9\pi(2-\sqrt2)\]tadah! (I have to go soon)
TuringTest
  • TuringTest
those bounds on the first integrals are switched a bit...
TuringTest
  • TuringTest
\[\int_{0}^{\frac\pi4}\int_{0}^{2\pi}\int_{0}^{3}\rho^2\sin\phi d\rho d\theta d\phi=9\int_{0}^{\frac\pi4}\int_{0}^{2\pi}\sin\phi d\theta d\phi\]\[=18\pi\int_{0}^{\frac\pi4}\sin\phi d\phi =-18\pi\cos\phi|_{0}^{\frac\pi4}\]\[=-18\pi(\frac{\sqrt2}2-1)=9\pi(2-\sqrt2)\]there we go
anonymous
  • anonymous
Genius, thank you!!!
TuringTest
  • TuringTest
I think you just made some algebra mistakes, your setup was good but thanks, and you're welcome :)

Looking for something else?

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