Calculus III: Calculating Volumes

[north nacelle]

I had to calculate the volume that is bounded by the following equations:
x^2+y^2+z^2 = 1 and x^2+y^2+z^2=16
z^2=x^2+y^2, x= 0 and y =0
and x,y,z >=0

Could someone please help me understand how to start with this stuff, I really have a hard time figuring out what region and what bounds

[azure minnow]

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[left spade]

the first two are spheres of radius 1 and 2 respectively

centered at 0

convert to spherical coordinates and determine the limits

[north nacelle]

yeah I know that I should convert to spherical coordinates, however I am stuck at knowing what region is being asked

[north nacelle]

yeah, sorry should be radius 4

[left spade]

this condition puts bounds to your limits of integration

what are they?

[north nacelle]

wdym what are they?

they ask you to find the region that is bounded by these two spheres, a cone, and planes and they must be all positive or equal to 0

[left spade]

but in spherical coordinates

you have 3 parameters, for which parameter values do you restrict to the given octant?

also draw pictures, always draw pictures for such problems

suppose you forget about the cone, how do you calculate volume between the two spheres for x,y,z nonnegative?

write out the integral in spherical coordinates

(dont forget the jacobian)

[north nacelle]

r,theta,delta

[north nacelle]

yeah, but drawing in 3D is pretty hard

[left spade]

sketch, draw projections

you dont need to be precise in your drawing

or study simpler examples before you tackle this one

[north nacelle]

the thing is I am having a hard time just finding out what region I am trying to find and the bounds are sometimes confusing

[left spade]

how about volume of a ball centered at 0 for x,y,z > 0 in spherical coordinates?

this is 1/8 volume of a ball, should be easy enough to figure which region this is

[azure minnow]

@north nacelle

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[north nacelle]

sorry I completely forgot

[north nacelle]

first octant I guess

[north nacelle]

hey do you mind if I ask something?

[left spade]

ask

[north nacelle]

like how do you know when to project it into some plane and do you draw stuff in 3D, because drawing stuff in 3D is pretty hard

[left spade]

i might draw sketches to get an idea of the behaviour of the limiting surfaces

the region is bounded between z=1 and z=4

then it's just a matter of figuring out what the cone looks like

[north nacelle]

yeah, so I know its two spheres a cone, and two planes. But then do you look at the xy plane or xz plane or yz plane, how do you determine that to find out the bounds

[north nacelle]

ah you don't draw it in 3D, I see

I mean yeah that makes it a ton easier to look at instead of some 3D drawing

[north nacelle]

what if I projected it into the xy plane?

[left spade]

then you are looking at it from the top

or bottom

[north nacelle]

and you did z-xy?

[left spade]

i look at it from the side

[north nacelle]

and why did you choose to do so?

or doesn't that matter?

[left spade]

it doesn't matter

your main task is to figure out what you are integrating and how the parameters change

whether you do it in cartesian form or spherical, cylindrical whatever you choose

[north nacelle]

Oh yeah

[left spade]

if you take top view this is what you see

see, you're much better at this

[north nacelle]

oh you don't see the line?

[north nacelle]

nah u are ;D

so well do you have then 2 option to choose from ( as in which region)

[left spade]

the labels dont make sense in this picture

[north nacelle]

because there are two regions you can cchoose from since they're both bounded

it's not on scale though

[left spade]

if you show the cone is going "up" then it should go up in the z direction

[north nacelle]

well what I did is just for every equation I was given set z= 0 and then draw these stuff

[left spade]

then you are seeing a projection onto xy plane

[north nacelle]

yeah

[left spade]

if you look at the cone from the top (or bottom) the only thing you see is a circle

not a triangle

if you look at it from the side then you see the V shape

[north nacelle]

oh because x^2+y^2 = z^2 into xy plane gives x^2 = -y^2 which cant be true , it is only true for x = 0

so it doesn't give me these two lines

[left spade]

let's say you put a bound z = 4

then you will see the large sphere and cone having the same projection

which is just a circle

[north nacelle]

yeah both are circles

ah I see

so it's better to do a side projection

because then you have the region that is bounded

[left spade]

both are useful

in spherical coordinates you pretty much have two parameters figured out now

[north nacelle]

the angle, theta and r

which is 1 < r < 4 and theta going from 0 to pi/2

[left spade]

the radial distance always goes from 1 to 4 and the polar angle from 0 to pi/2

correct

[north nacelle]

yeah!

[left spade]

so phi is definitely between 0 and pi/2 because you have to remain in the first octant

[north nacelle]

okay, just one sec

[left spade]

from which phi are you inside the cone?

[north nacelle]

there seem to be two different variants of spherical coordinates ( depends on how you take the angle phi) which is why I am a little bit confused on that one

since we have for z = r cos(phi) but on yt they have z = rsin(phi)

[left spade]

you mean from where you measure phi?

[north nacelle]

yeah

[left spade]

always from the z axis

[north nacelle]

we are doing it I think from the x-axis

yeah, not sure what azimuthal means

[left spade]

delta is in the inclination

but typically you measure delta from the z axis with delta in (0, pi)

[north nacelle]

yeah, they used a different from I guuess

so basically now I have to find out the angle phi, so the region obunded by everything

however, the region on the xy plane why do we cover it all, if the region needs to be also bounded by the cone

[left spade]

shouldn't be too hard, the answer's pretty much infront of you 😄

[left spade]

think about this picture where you are looking at the cone from the side

what do you think the angle is from z-axis to the boundary of the cone

[north nacelle]

oh no I meant

the angle theta

[left spade]

theta is already figured out

to stay in the first octant it can only be 0 to pi/2

[north nacelle]

yeah, but why though because we didn't figure out the cone boundary so would that not be causing troubles

I would've said from pi/4 to pi/2

but thats phi i guess

[left spade]

good!

now plug in the numbers and don't forget the jacobian

[north nacelle]

yeah, but just one question if you don't mind

[left spade]

ask

[north nacelle]

why isn't also theta going from pi/4 to pi/2 because isn't that the region we want

[north nacelle]

here we covered everything

ah but it's not the same actually

[left spade]

theta traces the points on the xy projection

[north nacelle]

that one is in the xy plane

[left spade]

you want to be precisely in the 1st quadrant

which is 0 to pi/2

[north nacelle]

yeah

thats why the xy plane is useful yup

[north nacelle]

yup approximately 9.66

now they also asked me to find it in cylinder coordinates, let me try that one

[north nacelle]

theta is the same 0 to pi/2

im just not sure with r

and you can also change x^2+y^2 to r^2

[left spade]

you have it sketched correctly

integrate in two parts

[north nacelle]

yes

[north nacelle]

so r is always from the origin to the surface, however now its not from the origin

I would just say its from 1 to 4 again

[left spade]

not with cylindrical

[north nacelle]

oh

[left spade]

you fix a value of z and you get a disc that contains all the points at that z

the disc is parallel to the xy plane

you want to know where this cut occurs

[north nacelle]

1/sqrt(2)

[left spade]

you are not integrating from z=1 to z=4 now

[north nacelle]

im not really seeing it why that is

[left spade]

your lower bound is either surface of sphere or the boundary of the cone

[north nacelle]

ah r is from the z-axis to the point?

[left spade]

mhm

if you fix z=2 for example so you're somewhere half way

it would be a straight line to the cone from z=2

[north nacelle]

yes, but so

[left spade]

that was correct

[north nacelle]

yeah, but so

[left spade]

see if your z is too small you are outside of the region

[north nacelle]

yeah

but if r starts at the intersection

don't you count that little region

now it goes from 1/sqrt(2) to 4

[left spade]

which little region

[north nacelle]

where the dotted line is

[left spade]

which one

[north nacelle]

the bottom

[left spade]

the lower bound is always the surface of the smaller sphere

[north nacelle]

yes, but the is the distance fromm the z-axis to that point, so if we're at z where the intersection happens

you get the thing outside the region

[left spade]

mhm, so at which distance does that occur?

[north nacelle]

x = 1/sqrt(2) is the intersection

[left spade]

what is the length of this dotted line

[north nacelle]

1/sqrt(2)

[left spade]

now write out the volume for this part in cylindrical form

[left spade]

cant be

[north nacelle]

that little region is confusing a bit

[left spade]

you always let r go from 1/rt2 to 4

so you're completely outside of the blue region

[stuck skiff]

hi

What is going on, in here?

[north nacelle]

hes helping.

[left spade]

multivariable calculus

[stuck skiff]

What exactly?

[left spade]

integration

[north nacelle]

you can see the question above

[left spade]

you can read the entire chat history

[north nacelle]

ohhh

from 0 to 1/sqrt2

[left spade]

better but still not true

[north nacelle]

hm

let me think

[left spade]

so you start z lower than 1

and also end z higher than 4

but your values of r depend on z

if we're just talking about the blue region

for now

[north nacelle]

yes

[left spade]

if z is between 1 and 4, then r is between 0 and 1/rt2

[north nacelle]

this?

[left spade]

but z can dip below 1 and go above 4 as well in the picture

[north nacelle]

yeah

why above 4?

we''re only stuck with the extra region at z < 1

[left spade]

uhh, this picture is not centered, I got confused

[north nacelle]

oh yeah sorry

but since z is bounded by the two

doesn't it account for it

so you just have r from 0 to 1/sqrt(2)

and then r going from 1/sqrt(2) to sqrt(8)

[left spade]

here you have equilateral right triangle with c=1, so the side length is 1/rt2 right?

meaning the lowest value of z is 1 - 1/rt2

but at that value of z, r only has one value, r=1/rt2

[north nacelle]

yes

[left spade]

so for z below 1, your possible values of r look like this

[north nacelle]

yeah, so the bounds for z accounts for that right?

if you had some other bound for z, you would maybe get that extra region from the circle but since the lower boundary is the sphere in this case

you get those all the possible r's without the other regions that you don't need ( if you set it up properly)

[left spade]

if z is above 1 you can integrate r from 0 to 1/rt2 for a long time

it's just below 1 and below 4 what you need to be careful with

[north nacelle]

but z here can be smaller than one

since you need that little region that is bounded by the circle and the line, that you showed with blue

[left spade]

mhm

so we use the relationship z^2 + r^2 = 1

whence r^2 = 1-z^2 and

$$\text{if } \frac{1}{\sqrt{2}} < z < 1, \text{then } \sqrt{1-z^2} < r < \frac{1}{\sqrt{2}}$$

so if you plug in the lowest value of z, you get only one possible value for r

[pastel panther]

aL

[left spade]

fixed typo

[north nacelle]

hmmm

This little thing is what I am not really understanding

[left spade]

to sum this up, if you want the volume of the green region in cylindrical form you do this

[north nacelle]

yes

[left spade]

$$\int _{1/\sqrt{2}}^1 \int _{0}^{\pi/2} \int _{\sqrt{1-z^2}}^{1/\sqrt{2}} |J|drd\phi dz + \int _1^{\sqrt{31/2}} \int _{0}^{\pi/2} \int _0^{1/\sqrt{2}} |J|drd\phi dz + \int _{\sqrt{31/2}}^4 \int _0^{\pi/2} \int _0^{\sqrt{16-r^2}}|J|drd \phi dz$$

[north nacelle]

uhh

[pastel panther]

aL

[left spade]

first term for below z=1

then up to a certain point r is always between 0 and 1/rt2

and then third term for z just below 4

[north nacelle]

lemme check

[left spade]

but remember, this is only volume for the green region

not the entire region

[north nacelle]

damn

but

[left spade]

for this remaining part you values of r always start from 1/rt2

[north nacelle]

let's say I already said that z has two boundaries, the small and big circle, doesn''t it account for it?

[left spade]

and are bounded by either the cone or the sphere

[left spade]

no

this is not how cylindrical coordinates work

you are not looking at points from the origin like in spherical case

you fix a value of z, let's say z= 1.5

and then you only see the points that are on the disc at that height

in spherical r from 1 to 4 is correct because it goes precisely from one boundary to the other

that's why it's useful to solve it in spherical form in the first place, because the boundaries themselves are spheres

[north nacelle]

yeah

ty man, you helped me a ton!

[left spade]

bedtime for me anyway

buenas noches

[north nacelle]

goodnightt!

[north nacelle]

Sorry for a lot of questions, but I am just trying to grasp it all. So if you wanna use cylindrical coordinates in the xy plane to find out the r bounds you can’t? You can only find it with the side projection because then you have your z-axis until your point? Or can you also use xy plane but in that case it’s from the origin?

[left spade]

there are no points on the xy plane in the given region

no matter the coordinate system you use, the question is always the same

what are the bounds such that all points are covered

[left spade]

the side projection helps you determine the z value of certain intersections

the cone intersects with the spheres, you need both z values

[north nacelle]

Yeah, I was just wondering because I was trying an exercise and well, I wanted to find the r bounds but I was in the xy plane so I wasn’t sure if you just apply cylindrical coordinates there

[left spade]

what is the exercise?

[north nacelle]

Give me a second to show you

I’ll also show you what I did

Find the volume that is bounded by these

so that purple region is what we need, we can already find r and theta right

theta is just between pi/3 and 2pi/3 which is pretty easy

but in this case I am not sure if we look at r now from the origin if we use cylindrical coordinates

for cylindrical coordinates it should be from the z-axis as you said

so should I look at a side projection?

[left spade]

this is top view, you only see this part

this tells you 0 < z < 1-r

[north nacelle]

Yes

Makes sense yeah I was thinking the same

[left spade]

i don't even know what that is tbh

,w plot (z+\sqrt{x^2+y^2})^2 = 1

[pastel panther]

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[left spade]

nvm, i think it's unit sphere

if x^2+y^2 = 1 that means you are at the boundary of the sphere, so z has no room to change

[north nacelle]

that is probably a cone

which is shifted by 1

because sqrt(x^2+y^2) = 1-z <=> x^2+y^2 = (1-z)^2

[left spade]

ah

good eye

a cone that opens downwards

[north nacelle]

yeah

[azure minnow]

@north nacelle

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[north nacelle]

+close

[azure minnow]

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