Question

use polar coordinates to find the volume of a sphere of radius a. Help, please

Answer 1

I set up theta from 0 to 2pi and r from 0 to a , the function is 2pi r^2 but the result is not right.:(

Answer 2

Everything always falls into place once the integral is set up.

Answer 3

@terenzreignz what do you mean?

Answer 4

Terence is epic.

Answer 5

Integrate the function.

Answer 6

^But to integrate a function, you need to have... A FUNCTION

Now tell me, @Hoa
What is the equation of a sphere of radius A, centred around the origin?

Answer 7

Sorry, radius a.

Answer 8

:')

Answer 9

2pir^2

Answer 10

sorry a^2

Answer 11

Hardly the equation of a sphere, @Hoa
But don't worry, that's why I'm here :)

The equation of a sphere centred around the origin, with radius a, is
\[\LARGE a^2=x^2+y^2+z^2\]

Answer 12

yeap

Answer 13

You'll notice it's similar to the equation of a circle right? In fact, we might as well put that up...
\[\LARGE r^2=x^2+y^2\]

Answer 14

Now, even though this isn't the correct approach, when converting from rectangular coordinates to cylindrical, you use this relation, as written above, right? ^
You need to change coordinate systems from
\[\LARGE (x,y,z) \rightarrow (r,\theta,z)\]
Fortunately, and clearly, z remains unchanged, so it's only a matter of the first two coordinates.

Answer 15

Conveniently, in the equation of the sphere with radius a, the part \(\large x^2+y^2\)
is already present.

\[\LARGE a^2=x^2+y^2+z^2\]

Answer 16

But first things first.. you've already set up the limits of the function, but now, we have to find THE function. It's almost always a function z = f(x,y).

Can you express z as a function of x and y? (Remember that a is constant).
@Hoa

Answer 17

x^2 +y^2 =a^2

Answer 18

^That is not a function. What I'm asking you to do is to isolate z, and express it as a function of x and y. Can you do that?

Answer 19

you mean z=sqr(a^2 -x^2 -y^2)?

Answer 20

Much better.

Answer 21

+-

Answer 22

\[\huge z = f(x,y) = \sqrt{a^2-(x^2+y^2)}\]

Answer 23

why don't we take - value?

Answer 24

No, we are not taking +/- values.
Just the positive values.

Answer 25

because the value of a volume must be a positive number?

Answer 26

We can do that, but that complicates things, don't you think? If we take only the positive values, we are only taking the volume of the upper hemisphere of the sphere.

In other words, only half the sphere. This will be rectified when we get to the integral.

We are only taking positive values because we need a FUNCTION, which, as you know, may not have more than one value, for any specific x-value or y-value.

Answer 27

and then *2?

Answer 28

That is correct. You catch on quick. ;)

Answer 29

So, we have a function. But we need it to be a function in polar coordinates...
How on earth do you change a function f(x,y) into a function g(r,θ) ?

Answer 30

by using x =rcos theta y = r sin theta?

Answer 31

Well, as I said earlier, the expression \(\large x^2 + y^2\) appears, quite conveniently, and may readily be converted to \(\large r^2\)

Answer 32

Well, you could do that, but you'll get the same result... after a little more crunching. ;)
Best keep things simple.

Answer 33

ok, so the function is sqr (a^2 -r^2)?

Answer 34

That is correct.

Answer 35

So, now, we integrate our function
\[\huge \iint\limits_R \sqrt{a^2-r^2} \ dA\]

Answer 36

You've defined the region over which we integrate, right?

Answer 37

yes, sir . everything done after set up the limits as I do above, right?

Answer 38

hey, hold on, how about one more r?

Answer 39

No calling me sir. ;)
Let's set up the limits...
\[\huge \int\limits_{\theta=0}^{2\pi}\int\limits_{r=0}^{a}\sqrt{a^2-r^2} \ dA\]

Answer 40

just a note,
\(\huge V=2\iint\limits_R \sqrt{a^2-r^2} \ dA \\ dA =r dr d \theta\)

Answer 41

Now, in polar coordinates, you can just take this to faith, unless you intend to take Real Analysis... but
\[\huge dA=dxdy=r\cdot drd\theta\]

Answer 42

@hartnn got it.

Answer 43

Thanks terenzeignz, thanks a lot everybody, I am ok from then.

Answer 44

So, the integral becomes
\[\huge \int\limits_{\theta=0}^{2\pi}\int\limits_{r=0}^{a}\ r \sqrt{a^2-r^2} \ drd\theta\ \]

Answer 45

bingo.

Answer 46

You can do it from here?

Answer 47

sure.

Answer 48

If you say so. Good Hunting :)

Answer 49

do you want to check?

Answer 50

let u = inside sqr, change the limit. and... done, right?

Answer 51

I'll leave you to your methods. :)

Answer 52

I do shortcut, but trust me, I can do it. thanks a lot.

Answer 53

Though, you're welcome to show your process.

Answer 54

under those pressure?

Answer 55

Pressure is an illusion...
They are not there to intimidate you, they are there... to cheer you on.

Get inspired.

And did I mention... don't call me sir? Good :)

Answer 56

ok, thanks friend!!!

Answer 57

No problem :)

Answer 58

to me, u = a^2 -r^2 ---> du = -2 r dr , substitute into the function to get -1/2sqr (u) du

Answer 59

change the limit , and take int as usual, it's quite tedious but it is easy

Answer 60

Stand by while I attempt to locate a sheet of paper and a pencil.

Answer 61

at the end up, the volume of a sphere must be 4/3 pi r^3

Answer 62

a, not r.

Answer 63

yes. sorry. a^3

Answer 64

ha ha.. you got 5 medals from my question, cheeerr !!!

Answer 65

Isn't that just great :)
Here, take mine ...

Answer 66

haha...ridiculous us!!! cheer up each other!!! ok
anyways, I appreciate what you do for me.
take care

Answer 67

Could you restate your new question?

Answer 68

find the volume of the solid above z= sqr(x^2 +y^2) and below the sphere x^2 +y^2+z^2 =1. I confused

Answer 69

hehe... I am safe now, no one monitor, no pressure.

Answer 70

You could do this normally, with rectangular coordinates, but I think polar is the way to go here, considering the ease of integration. Do you agree with me on that count? :)

Answer 71

yeap

Answer 72

Well, remember in 2d, when you have two functions and are asked to find the area between them

You simply integrate the function f(x) - g(x) right?