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

- anonymous

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- schrodinger

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- anonymous

@ganeshie8

- ganeshie8

Try
\(\rho : ~0\to 1 \)
\(\theta : ~0\to 2\pi \)
\(\phi : ~\frac{\pi}{3}\to \frac{\pi}{2}\)

- anonymous

as my limits for triple integration?

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## More answers

- ganeshie8

Yes

- anonymous

p is dx θ is dy ϕ is dz?

- anonymous

whats the functions im integrating?

- ganeshie8

Hmm @Loser66 I didn't get your question..

- ganeshie8

@jyar what do you know about spherical coordinates ?

- anonymous

there related to the cartesian coordinates and its in 3d points (r, θ , ϕ )

- ganeshie8

what do \(\rho, \theta\) and \(\phi\) represent ?

- anonymous

angles projected from the xyz plane

- Loser66

.

- anonymous

p^2=x^2+y^2+z^2

- ganeshie8

|dw:1436648757792:dw|

- 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

oh okay i understand that

- 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

Also \(\theta\) varies from 0 to 2pi is also trivial

- ganeshie8

you need to do some work to figure out the bounds for \(\phi\)

- anonymous

is θ always 0 to 2pi?

- 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

2 full revolution would be 4pi?

- ganeshie8

You never want to do 2 full revolutions as that might duplicate the volume

- ganeshie8

btw you're correct about 4pi being two full revolutions

- anonymous

oh okay so when finding the limits of ϕ what would i do

- ganeshie8

the part of sphere between z=1/2 and z=0 looks like below?
|dw:1436649362434:dw|

- anonymous

yeah i see that

- ganeshie8

z=0 represents the xy plane,
whats the angle \(\phi\) for xy plane ?

- ganeshie8

Remember, \(\phi\) is the angle from z axis

- anonymous

pi/3 to 0?

- ganeshie8

z axis is perpendicular to xy plane, yes ?

- anonymous

yes

- ganeshie8

so whats the angle between xy plane and positive z axis ?

- anonymous

its looks 60 degrees

- anonymous

its not that ?

- ganeshie8

Easy, xy plane makes 90 degrees with the positive z axis.

- ganeshie8

|dw:1436649826870:dw|

- ganeshie8

that means pi/2 is the ending angle
lets find the starting angle

- anonymous

its not under pi.2 its a little further away...so 17pi/12?

- ganeshie8

|dw:1436649896094:dw|

- ganeshie8

see if you can find \(\theta\) in that right triangle

- ganeshie8

Honestly I have no idea how you got 17pi/12

- anonymous

60 i got

- ganeshie8

Yes starting angle is 60 degrees, which is same as pi/3

- anonymous

yes

- 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

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

ok! the first integrations foes with p^2dp, second sinϕ dϕ, third dθ?

- ganeshie8

It doesn't matter,
only the bounds and the differentials need to agree

- ganeshie8

Now that we have moved to spherical,
\(\large \color{blue}{\rho^2 \sin\phi}\) is your integrand here

- anonymous

ok i got that...could we do the first integration

- 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

that red part

- anonymous

the 1 intregrated becomes just ϕ?

- ganeshie8

integrate below and plug it in
\[\large \color{red}{\int\limits_{\pi/3}^{\pi/2}~1~\rho^2\sin\phi~d\phi}\]

- ganeshie8

thats same as
\[\large \color{red}{\int\limits_{\pi/3}^{\pi/2} \rho^2\sin\phi~d\phi}\]

- 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

i got (-1/162) without pulling the p^2 out

- anonymous

pulling the p^2 i get (1/2)p^2

- anonymous

(-1/2)

- 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

ok yes...for the next integration i \[\int\limits_{0}^{2\pi} (1/2)p^2 dp\]

- 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

Nope, next we work :
\[\large \int\limits_{0}^{2\pi} (1/2)p^2 d\color{red}{\theta}\]

- ganeshie8

keepin mind \(0\to 2\pi\) are bounds for \(\theta\),
not \(\rho\)

- anonymous

oh okay so those bounds also how the order of work....
for the next intregation i got (4pi/3)

- ganeshie8

\[\large \int\limits_{0}^{2\pi} (1/2)p^2 d\color{red}{\theta} = ?\]

- anonymous

i pulled out the (1/2) and the inside i got (2pi^3/3) = (8pi/3)= (1/2)(8pi/3)?

- ganeshie8

looks wrong,
you can pull out entire thing, everythng is constant there

- 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

(pi) p^2?

- ganeshie8

Yes, plug that in the volume integral

- anonymous

what if i didnt pull everything out as a constant and instead integrated would that be all wrong?

- 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

pi/3?

- ganeshie8

Yep!

- 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

can you show me how exactly are you "integrating" everything ?

- anonymous

\[\int\limits_{0}^{2\pi}(1/2)p^2 d\]

- anonymous

wait i think i got it !
wow thank you soo much! so helpful

- ganeshie8

d what ?

- anonymous

d theta

- ganeshie8

feel free to ask if you have any questions :)

- anonymous

will you be online for rest of today?

- ganeshie8

il be around for 1 hour or so, feel free to tag me in ur questions

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