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

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

- 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?

Looking for something else?

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

## 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

Looking for something else?

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