Question

I'm having problems with this integral. "Use polar coordinates to find the volume of the given solid - under the cone z=sqrt(x^2+y^2) and above the disk x^2+y^2<=4"

Answer 1

We should use cylindrical coordinates for this integral. These are similar to polar coordinates with the addition of z coordinate. z goes from 0 to r, r goes from 0 to 2 and theta, the angle goes from 0 to 2*pi so the integral will be:
\[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{0}^{r}r dz dr d \theta\]
notice how you get an extra r in your integral when you go from regular coordinates to cylindrical coordinates.

Answer 2

Well, since the problem says to use polar coordinates, I think we should try to do it that way. Some professors take off for using a different method.

Answer 3

First thing to do is try to picture it. I'm working on a visual representation now.

Answer 4

Let's check with trunsaid, they may not even be at triple integrals yet.

Answer 5

I tried doing polar coordinates and I'm getting

\[\int\limits_{0}^{2\pi}\int\limits_{r}^{2}rdrd \theta\]

Answer 6

I don't see how you can determine the volume of three dimensional object with a two dimensional coordinate system.

Answer 7

Yes, you can but there would be a function in integrand

Answer 8

Refer to picture attached.

Answer 9

Thomas, polar coordinates are 3 dimensional. There is a second angle.

Answer 10

smoothmath, I thought that was spherical coordinates that has two angles?

Answer 11

Yes, Phi and theta

Answer 12

Well, spherical is an extension of polar.

Answer 13

Okay, so theta is the angle with the x axis going around the z axis, okay? And Phi we'll call the angle with the x axis going around the y axis.

Answer 14

With polar coordinates:
\[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}r*rdrd \theta\]

this is the same as when you use cilindrical coordinates. There are now two r's in the inegrand one from the transformation and on for the height.

Answer 15

Refer to picture for angles.

Answer 16

Following, Trung?

Answer 17

spherical coordinates are useless here, because you have cone and not a sphere of some kind.

Answer 18

You're wrong, Thomas, and if you follow along, you'll see that.

Answer 19

They actually make it really simple.

Answer 20

@smoothman go for it

Answer 21

theta goes from 0 to 2pi, obviously.

Answer 22

Phi goes from 0 to arctan(2/4)

Answer 23

And then r depends on phi. So it goes from 0 to 4/cos(phi)

Answer 24

See it?

Answer 25

\[\int\limits \int\limits \left(\sqrt{x^2+y^2}-\left(x^2+y^2-4\right)\right)dydx\]
\[\int _0^{2\pi }\int _0^2\left(r-\left(r^2-4\right)\right)rdrd\theta \]

Answer 26

I have a feeling trung is just gone, but hey.

Answer 27

@smooth math, yea i understand how you did that

Answer 28

@smoothman no, you might be correct, but it sure isn't making it easier.

Answer 29

Well do you see the angles I drew? Theta goes from 0 to 2pi, right?

Answer 30

i agree

Answer 31

@imranmeah91, subtracting the two equations just took care of the change in R distance as you're integrating?

Answer 32

Next we want to see the limits for phi. It starts at 0. Where does it end? The cone forms a right triangle with adjacent side 4 and opposite side length 2.

Answer 33

So the angle for that is arctan(2/4). So phi goes from 0 to arctan(2/4)

Answer 34

That's what we need to find volume, I believe

Answer 35

This thread is getting messy, isn't the adjacent side 2?

Answer 36

Yeah, to introduce spherical is to further complicate the problem and you are not following the teacher's instructions, which is to use polar coordinates. (Even if you say spherical is just an extension...) I think trunsaid should just follow imranmeah's example.

Answer 37

Ah yes. Definitely. My mistake.

Answer 38

thank you all for the help though. all of you made this whole concept of multiple integrals easier to understand

Answer 39

Hmmm. You're right. You can do it in pure polar if you work with a third variable, z.

Answer 40

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}r*rdrd \theta\] I still think this is correct

Answer 41

z from 0 to 2, then theta from 0 to 2pi, then r from z to 2.

Answer 42

Magic.

Answer 43

My bad, guys. I was complicating things. I'm just too stubborn.

Answer 44

honestly though using cylindrical coordinates made this much easier for me but I'm pretty sure my professor wanted me to use polar

Answer 45

This is the front view

Answer 46

Last picture, I promise. This should clarify things.

Answer 47

I don't think yours quite does it, Imran. r isn't purely from 0 to 2.

Answer 48

I don't see how you get that integrand imran.

Answer 49

It's a three dimensional region, so you need a third variable, z. r depends on z.

Answer 50

Thomas, does my approach make sense to you?

Answer 51

z=r, so why not make the integrand r*r?

Answer 52

@smoothman, sort of, but as I've said, z=r, so just put r in the integrand

Answer 53

Thomas9 is right
the height is \[\sqrt{x^2+y^2}\]=r
add up the heights at all radius and angle
\[\int _0^{2\pi }\int _0^2r\text{ }rdrd\theta \]

Answer 54

I don't see why that wouldn't work.

Answer 55

me neither

Answer 56

doesn't Thomas9's way only make sense if the radius is constant?

Answer 57

No, I think it works fine because you integrate for r from 0 to 2. You aren't treating it as a constant.

Answer 58

My way I think is a little more intuitive, but I think his is right too.

Answer 59

And only a double integral while mine is a triple, so that's cool.

Answer 60

Indeed, r is not constant.

Answer 61

I still think the spherical triple integral is a particularly elegant solution. =D