# Trouble converting Cartesian to Polar

longbusy
Well, it is not really hard to convert them. My main problem is thinking in Polar coordinates. Cartesian coordinates are really easy to think about for me (after how many years of experience) but then I get to Calc 3 and I hit a brick wall. Does anyone have some insight on how to get past this problem?

-Jeremy

i.e.

x^2 + y^2 = 9 gives a cylinder of radius 3 in 3-D

the same equation is r = 3, right? That's pretty simple, but once it gets more complicated Boom, I'm lost.

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Ellipse (center at origin, axes of ellipse along coordinate axes)

(x^2/a^2) + (y^2/b^2) = 1

Where a and b are the two semi-axes.

Hyperbola (similar conditions)

(x^2/a^2) - (y^2/b^2) = 1

Parabola (vertex at origin, axis along y axis opening up)

y = ax^2

Originally posted by longbusy
Does anyone have some insight on how to get past this problem?

Just lots of practice to get familiar with them, just keep tredging through it and do your homework and after a while it will all start making sense. In fact for many problems the polar, or spherical coordinate systems will be come your coordinates of choice becuase they make many things much much simpler. Have you gotten to the change of variables topic yet in calc 3? If not, you'll see what I mean once you get there, many integrals are much easier to perform in sperical or cylindrical coordinates rather then in cartesian.

longbusy
Actually, the substitution part is what we're working on right now. That is why I asked the question. It's killing me. :) I'll just keep trudging. Thanks!

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Well, substitution can mean different things... which operation gives you trouble when substituting?

Can you apply the chain rule for differentiation when substitutions are involved? e.g.:

Letting @ be the symbol for partial differentiation:

@f/@r = @f/@x * @x/@r + @f/@y * @y/@r
= cos &theta * @f/@x + sin &theta * @f/@y

Did you understand substitution in integrals in calc 2? i.e.:

&int sqrt(1 - x^2) dx = &int sqrt(1 - (sin &theta)^2) cos &theta d&theta = ...

and by understand I mean that you could replace all occurances of one variable with the other (x -> sin &theta) in the integral, compute the equation for replacing the differential (dx = cos &theta d&theta), and you could compute the new bounds for your integral if it happened to be a definite integral.

Do you understand how to perform substitution in a multiple integral, meaning you can both replace the variables and the differentials? E.G.

via Jacobian:
x = r cos &theta, y = r sin &theta
dx dy = J(x, y / r, &theta) dr d&theta
Code:
= det /@x/@r   @x/@&theta\ dr d&theta
\@y/@r   @y/@&theta/
= det /cos &theta   -r sin &theta\ dr d&theta
\sin &theta    r cos &theta/
= (r (cos &theta)^2 + r (sin &theta)^2) dr d&theta
= r dr d&theta

or via algebra: (I'm lazy so I won't compute things, just label 'em with capital variables)
dx dy = (@x/@r dr + @x/@&theta d&theta) * (@y/@r dr + @y/@&theta d&theta)
= A dr dr + B dr d&theta + C d&theta dr + D d&theta d&theta
= 0 + B dr d&theta - C dr d&theta + 0
= (B - C) dr d&theta = r dr d&theta

Do you understand how to recompute the limits of integration in a multiple integral, or equivalently being able to translate a description of a region in one set of coordinates into another set of coordinates? e.g. going from:

x^2 + y^2 <= 9

to

r <= 3 and 0 <= &theta <= 2&pi

Hurkyl

longbusy
Hurkyl, thanks, I am working through some topics you listed now. I did well in Calc 2 with the substitution. It is just stuff like the following:

"Use a double integral in polar coordinates to find the volume of the solid that is described."

x^2 + y^2 + z^2 = 9 (inside)
x^2 + y^2 = 1 (outside)

The surface is a circle with a cylinder through it (duh). I know I want the area from the cylinder wall out to the edge of the circle. I know that the lower bound for r will be 1. But the upper bound of r and the bounds for theta are kind of beyond me at the moment.

Though, I am working these a little easier now.

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So the description of your volume is:

x^2 + y^2 > 1
x^2 + y^2 + z^2 < 9

Are you having trouble simple converting these equations into polar form, or can you get that far and just have trouble figuring out what the bounds of integration are?

Hurkyl

longbusy
I am mainly having trouble figuring out the bounds. I can draw the pictures and see them. I cannot figure out the bounds. That is what is holding me up.

arcnets
Originally posted by longbusy "Use a double integral in polar coordinates to find the volume of the solid that is described."

x^2 + y^2 + z^2 = 9 (inside)
x^2 + y^2 = 1 (outside)

Hmmm, OK, lemme try.

x = r cos[psi] cos[the]
y = r sin[psi] cos[the]
z = r sin[the]

I. r^2 < 9
II. r^2 cos^2[the] > 1, or r^2 > 1/cos^2[the]

dV = r^2 cos [the] dr d[the] d[psi]
V = Int [0 ; 2[pi]] Int [-[pi]/2 ; [pi]/2] cos [the] Int [1/cos^2[the] ; 9] r^2 dr d[the] d[psi]

OK so far?

Last edited:
arcnets
Correction.

My Theta limits are wrong. We've got to keep 1/cos^2[the] smaller than 9, which means
1/cos^2[the] < 9
cos^2[the] > 1/9
|[the]| < arccos 1/3

So, should read Int [-arccos 1/3 ; arccos 1/3] ... d[the]

longbusy
Well, it's in polar coordinates, not cylindrical... But I assume those are correct. What I calculated:

r^2 = 9 - 9cos^2(theta)
-or-
r = 3 - 3cos(theta)

Now, the book claims that the bounds are:

8Int[0, pi/2]Int[1,3]...(dr)(d(theta))

I understand the first bounds (eight quadrants for the full circle), but why is it Int[1,3] for r?

Staff Emeritus
Gold Member
There are only 4 quadrants in the full circle (I imagine the extra factor of 2 was pulled from inside the integral to outside).

As for the bounds on r, go back to the inequalities I listed:

x^2 + y^2 > 1 (outside the cylinder)
x^2 + y^2 + z^2 < 9 (inside the sphere)

Convert those to cylindrical coordinates (cylindrical coordinates are the same thing as polar coordinates, just with z as a third coordinate) and you might see where the bounds on r come from! (If not, show me what you got as the result from converting both equations and I'll go from there)

Hurkyl

longbusy
I meant octants, not quadrants. Sorry! That explains the 8.

In cyclindrical coordinates, I got the following for x^2 + y^2 + z^2 = 9:

r^2 = 9 - 9cos^2(theta)
-or-
r = 3 - 3cos(theta)

So, r goes from 1 to 3 - 3cos(theta) right?

Thanks a lot for your help, I know I am being a bit dense.

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Gold Member
Where are you getting cos theta? z should just remain z when you do the conversion

Hurkyl

longbusy
Mixing up Spherical with Cylindrical for some reason. So it should remian r^2 + z^2 = 9?

Staff Emeritus
Gold Member
Right, it should remain r^2 + z^2 < 9 (don't forget the inequality; it specifies you care about the interior of the sphere rather than just its surface. More complicated problems can be made a little easier if you don't forget this)

From this inequality, you can find the largest possible value of r (which is 3 when z = 0), and then once you have your bounds on r you can solve for z to get the bounds on it.

Hurkyl

longbusy
Thanks! I am working on a lot more of these problems now. Better safe than sorry. I just cannot understand why only Odd answers are given in supplemental answer books. If professors do collect homework, work has to be shown... Just doesn't make sense. :)

Thanks again!

ObsessiveMathsFreak
one of these should be a symbol for partial differencation

delta &delta;
Delta &Delta;

actually it's this one [pard]