anonymous
  • anonymous
MEDAL***Please help use spherical coordinates to find the volume cut out from the sphere x^2+y^2+z^2=1 by the planes z=1/2 and z=0
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@ganeshie8
ganeshie8
  • ganeshie8
Try \(\rho : ~0\to 1 \) \(\theta : ~0\to 2\pi \) \(\phi : ~\frac{\pi}{3}\to \frac{\pi}{2}\)
anonymous
  • anonymous
as my limits for triple integration?

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ganeshie8
  • ganeshie8
Yes
anonymous
  • anonymous
p is dx θ is dy ϕ is dz?
anonymous
  • anonymous
whats the functions im integrating?
ganeshie8
  • ganeshie8
Hmm @Loser66 I didn't get your question..
ganeshie8
  • ganeshie8
@jyar what do you know about spherical coordinates ?
anonymous
  • anonymous
there related to the cartesian coordinates and its in 3d points (r, θ , ϕ )
ganeshie8
  • ganeshie8
what do \(\rho, \theta\) and \(\phi\) represent ?
anonymous
  • anonymous
angles projected from the xyz plane
Loser66
  • Loser66
.
anonymous
  • anonymous
p^2=x^2+y^2+z^2
ganeshie8
  • ganeshie8
|dw:1436648757792:dw|
ganeshie8
  • ganeshie8
For any point \((\rho, \theta, \phi)\) in space, \(\rho\) is the "distance" from origin \(\theta\) is the angle in xy plane with the positive x axis \(\phi\) is the angle with positive z axis
anonymous
  • anonymous
oh okay i understand that
ganeshie8
  • ganeshie8
For our specific problem, it is easy to see that \(\rho\) varies from 0 to 1 because the radius of sphere is 1
ganeshie8
  • ganeshie8
Also \(\theta\) varies from 0 to 2pi is also trivial
ganeshie8
  • ganeshie8
you need to do some work to figure out the bounds for \(\phi\)
anonymous
  • anonymous
is θ always 0 to 2pi?
ganeshie8
  • ganeshie8
It depends, from the diagram you can see that \(\theta\) is the angle in xy plane. \(\theta ~: ~0\to 2\pi\) means this angle is swept one full revolution
anonymous
  • anonymous
2 full revolution would be 4pi?
ganeshie8
  • ganeshie8
You never want to do 2 full revolutions as that might duplicate the volume
ganeshie8
  • ganeshie8
btw you're correct about 4pi being two full revolutions
anonymous
  • anonymous
oh okay so when finding the limits of ϕ what would i do
ganeshie8
  • ganeshie8
the part of sphere between z=1/2 and z=0 looks like below? |dw:1436649362434:dw|
anonymous
  • anonymous
yeah i see that
ganeshie8
  • ganeshie8
z=0 represents the xy plane, whats the angle \(\phi\) for xy plane ?
ganeshie8
  • ganeshie8
Remember, \(\phi\) is the angle from z axis
anonymous
  • anonymous
pi/3 to 0?
ganeshie8
  • ganeshie8
z axis is perpendicular to xy plane, yes ?
anonymous
  • anonymous
yes
ganeshie8
  • ganeshie8
so whats the angle between xy plane and positive z axis ?
anonymous
  • anonymous
its looks 60 degrees
anonymous
  • anonymous
its not that ?
ganeshie8
  • ganeshie8
Easy, xy plane makes 90 degrees with the positive z axis.
ganeshie8
  • ganeshie8
|dw:1436649826870:dw|
ganeshie8
  • ganeshie8
that means pi/2 is the ending angle lets find the starting angle
anonymous
  • anonymous
its not under pi.2 its a little further away...so 17pi/12?
ganeshie8
  • ganeshie8
|dw:1436649896094:dw|
ganeshie8
  • ganeshie8
see if you can find \(\theta\) in that right triangle
ganeshie8
  • ganeshie8
Honestly I have no idea how you got 17pi/12
anonymous
  • anonymous
60 i got
ganeshie8
  • ganeshie8
Yes starting angle is 60 degrees, which is same as pi/3
anonymous
  • anonymous
yes
ganeshie8
  • ganeshie8
so do the bounds make sense ? \(\rho : ~0\to 1 \) \(\theta : ~0\to 2\pi \) \(\phi : ~\frac{\pi}{3}\to \frac{\pi}{2}\)
ganeshie8
  • ganeshie8
The volume is given by \[\large V = \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \int\limits_{\pi/3}^{\pi/2}~1~\color{blue}{\rho^2\sin\phi~d\phi~d\theta~d\rho}\]
anonymous
  • anonymous
ok! the first integrations foes with p^2dp, second sinϕ dϕ, third dθ?
ganeshie8
  • ganeshie8
It doesn't matter, only the bounds and the differentials need to agree
ganeshie8
  • ganeshie8
Now that we have moved to spherical, \(\large \color{blue}{\rho^2 \sin\phi}\) is your integrand here
anonymous
  • anonymous
ok i got that...could we do the first integration
ganeshie8
  • ganeshie8
start by working the inner integral : \[\large V = \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \color{red}{\int\limits_{\pi/3}^{\pi/2}~1~\rho^2\sin\phi~d\phi}~d\theta~d\rho\]
ganeshie8
  • ganeshie8
that red part
anonymous
  • anonymous
the 1 intregrated becomes just ϕ?
ganeshie8
  • ganeshie8
integrate below and plug it in \[\large \color{red}{\int\limits_{\pi/3}^{\pi/2}~1~\rho^2\sin\phi~d\phi}\]
ganeshie8
  • ganeshie8
thats same as \[\large \color{red}{\int\limits_{\pi/3}^{\pi/2} \rho^2\sin\phi~d\phi}\]
ganeshie8
  • ganeshie8
you can pull out \(\color{red}{\rho^2}\) because it is constant with respect to \(\phi\) : \[\large \color{red}{\rho^2\int\limits_{\pi/3}^{\pi/2} \sin\phi~d\phi}\]
anonymous
  • anonymous
i got (-1/162) without pulling the p^2 out
anonymous
  • anonymous
pulling the p^2 i get (1/2)p^2
anonymous
  • anonymous
(-1/2)
ganeshie8
  • ganeshie8
\[\large \color{red}{\rho^2\int\limits_{\pi/3}^{\pi/2} \sin\phi~d\phi} = \color{red}{\rho^2 (-\cos\phi)~ \Bigg|_{\pi/3}^{\pi/2} } = \color{red}{\rho^2 [-(0-\frac{1}{2})]} = \color{red}{\frac{1}{2}\rho^2}\]
anonymous
  • anonymous
ok yes...for the next integration i \[\int\limits_{0}^{2\pi} (1/2)p^2 dp\]
ganeshie8
  • ganeshie8
plugging that in the volume integral we get \[\large V = \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \color{red}{\int\limits_{\pi/3}^{\pi/2}~1~\rho^2\sin\phi~d\phi}~d\theta~d\rho = \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \color{red}{ \frac{1}{2}\rho^2}~d\theta~d\rho\]
ganeshie8
  • ganeshie8
Nope, next we work : \[\large \int\limits_{0}^{2\pi} (1/2)p^2 d\color{red}{\theta}\]
ganeshie8
  • ganeshie8
keepin mind \(0\to 2\pi\) are bounds for \(\theta\), not \(\rho\)
anonymous
  • anonymous
oh okay so those bounds also how the order of work.... for the next intregation i got (4pi/3)
ganeshie8
  • ganeshie8
\[\large \int\limits_{0}^{2\pi} (1/2)p^2 d\color{red}{\theta} = ?\]
anonymous
  • anonymous
i pulled out the (1/2) and the inside i got (2pi^3/3) = (8pi/3)= (1/2)(8pi/3)?
ganeshie8
  • ganeshie8
looks wrong, you can pull out entire thing, everythng is constant there
ganeshie8
  • ganeshie8
\[\large \int\limits_{0}^{2\pi} (1/2)p^2 d\color{red}{\theta} = (1/2)p^2\int\limits_{0}^{2\pi} 1 d\color{red}{\theta} = ? \]
anonymous
  • anonymous
(pi) p^2?
ganeshie8
  • ganeshie8
Yes, plug that in the volume integral
anonymous
  • anonymous
what if i didnt pull everything out as a constant and instead integrated would that be all wrong?
ganeshie8
  • ganeshie8
plugging that in the volume integral we get \[\begin{align}\large V &= \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \color{red}{\int\limits_{\pi/3}^{\pi/2}~1~\rho^2\sin\phi~d\phi}~d\theta~d\rho = \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \color{red}{ \frac{1}{2}\rho^2}~d\theta~d\rho\\~\\ &=\int\limits_{0}^1 ~\pi \rho^2 ~d\rho\\~\\ &= ? \end{align}\]
anonymous
  • anonymous
pi/3?
ganeshie8
  • ganeshie8
Yep!
anonymous
  • anonymous
regarding my question for the second integration if i didnt pull everything out as a constant and intregrated evrthhing instead would that be wrong?
ganeshie8
  • ganeshie8
can you show me how exactly are you "integrating" everything ?
anonymous
  • anonymous
\[\int\limits_{0}^{2\pi}(1/2)p^2 d\]
anonymous
  • anonymous
wait i think i got it ! wow thank you soo much! so helpful
ganeshie8
  • ganeshie8
d what ?
anonymous
  • anonymous
d theta
ganeshie8
  • ganeshie8
feel free to ask if you have any questions :)
anonymous
  • anonymous
will you be online for rest of today?
ganeshie8
  • ganeshie8
il be around for 1 hour or so, feel free to tag me in ur questions

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