### Author Topic: Question: Using an integral and Sine to calculate the volume of a sphere  (Read 637 times)

#### Nicepoint

I tried to calculate the volume of a sphere with a radius of 5 with this:

(yes I noticed the 2 in the front is useless.)
which gave me 123.37006...

But when I compared it to the trusted V = (4/3)πr3
It gave me 523.59878...

Any of you know what I did wrong?

#### EcstaticEggplant

so you're trying to find the volume of revolution for half of the sine curve? i don't think you can just use it like that to represent a circle/sphere. without getting into polar coordinates you could just use the equation of a circle (x^2 + y^2 = r^2) and integrate pi(y^2)dx like π∫(r^2 - x^2) dx with the limits r and -r (replacing all r's with 5) to get the volume of revolution for a circle, giving you the volume for a sphere.

#### Nicepoint

Well, since Sine is the y-coordinate of the point on a unit circle, I thought that by multiplying it with the radius, would get me the y-coordinate on the actual circle and then integrating all of the y-coordinates on the 0 to pi/2. That would give me the area of a quarter of the circle, which I would multiply by two to extend it to the other side of the y-axis and rotate around the x-axis for the 3d-sphere.

I know the (x2 + y2 = r2) thing, but I thought I could also do it through the Sine function.

#### TableSalt

what language you guys speakin in here

#### .:FancyPants:.

generally when you learn about calculating anything involving a sphere with integrals, you first learn about cylindrical coordinates (r, θ, z) and spherical coordinates (ρ, θ, φ). these are just different ways of drawing up lines in 3D space that are like the cartesian coordinates (x, y, z) you already know.

the reason why you learn about these coordinate systems? because, among many other things, it makes calculating the volume of a sphere with integrals way nicer. we're talking about a volume, which is three-dimensional, so typically there's some way to write that as a triple integral. and in fact, you can generally write any calculation for two-dimensional area as a double integral, so that makes sense.

so here's what the equation ends up looking like when you want to calculate the area of a sphere with cartesian coordinates:

this is a huge pain. cylindrical coordinates, which would substitute slightly different terms into the x and y variables, will actually simplify this triple integral (although when you get into a calculus 3 class, these substitutions will make a lot more sense):

and spherical coordinates, which were basically born to do this, make it look like this:

(pictures shamelessly stolen from this pdf)

these equations can be expanded and compressed a thousand different ways to look different but actually still mean the same thing. in fact, all three of these equations can be dumbed down to V = (4/3)πr3 if you know what you're simplifying.

but your intuition is definitely right: in the most elegant way, a sine is involved in calculating the volume of a sphere.

#### Tactical Nuke

yeah you're not going to get volume from a single integral, it has to be at least double

#### Nicepoint

It's a sphere, I really don't need to do a double or triple integral when I can just spin it around the x-axis with:
π * ∫(F(x))2dx

All I need for it is just an integral of one quarter of the circle.
(I'm about to leave high school so I wouldn't know about double or triple integrals anyways)

#### St. Francis Xavier

It's a sphere, I really don't need to do a double or triple integral when I can just spin it around the x-axis with:
π * ∫(F(x))2dx

All I need for it is just an integral of one quarter of the circle.
(I'm about to leave high school so I wouldn't know about double or triple integrals anyways)

You'll probably like this then http://tutorial.math.lamar.edu/Classes/CalcIII/TISphericalCoords.aspx

#### Kearn

doing circular integrations in cartesian coordinates is kinda hellish, you need trig identities and substitutions to do it reasonably if you want to try using trig functions

using cylindrical or spherical coordinates is way easier, as previously mentioned

i think the way you'd have to do it is integrate an entire semicircle and rotate that around whatever axis you choose 360 degrees, but i think you'd still be integrating something obnoxious

the main problem is that you need something with a constant radius, so if you're just trying to integrate sin in cartesian you're going to wind up with a lopsided abomination

integrating y = sqrt(r^2-x^2) from 0 to pi and then rotating that around the x axis is a way easier method
« Last Edit: March 24, 2018, 03:36:03 PM by Kearn »