- #1

longbusy

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Thanks in advance!

-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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- Thread starter longbusy
- Start date

- #1

longbusy

- 19

- 0

Thanks in advance!

-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.

- #2

selfAdjoint

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(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

- #3

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.

- #4

longbusy

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- #5

Hurkyl

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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 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

- #6

longbusy

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"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.

- #7

Hurkyl

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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

- #8

longbusy

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- #9

arcnets

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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:

- #10

arcnets

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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]

- #11

longbusy

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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?

Thanks for all your help!

- #12

Hurkyl

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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

- #13

longbusy

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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.

- #14

Hurkyl

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

Hurkyl

Hurkyl

- #15

longbusy

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

- #16

Hurkyl

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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

- #17

longbusy

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Thanks again!

- #18

ObsessiveMathsFreak

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one of these should be a symbol for partial differencation

delta δ

Delta Δ

delta δ

Delta Δ

- #19

actually it's this one [pard]

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