Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Setting up polar-integrals

  1. Apr 6, 2009 #1

    I somewhat see the explanation but I need a conversation to clear the fog.

    Part of a larger problem is this integral:

    Find the volume of a cone [tex] r = 3sin(theta) [/tex] inside the sphere [tex] x^2 +y^2 + z^2 = 9 [/tex]

    \int \int \int r \,dz dr dtheta

    My limits:
    z goes from 0 to [tex] \sqrt{9-r^2}[/tex] with the integral multiplied by “2” to handle the +/- side of the square-root

    r goes from 0 to 3 sin(theta)

    and initially I had theta going from 0 to 2pi.

    First integration:

    \int \int r\sqrt{9-r^2}\, dr dtheta

    r from 0 to 3sin(theta)
    theta from 0 to 2pi – again initially – I reasoned that this is not correct

    Second integration:

    -\frac{2}{3} \int (9-r^2)^{2/3}\, dtheta

    r from 0 to sin(theta)

    -\frac{2}{3} \int (9-9sin(theta)^2)^{2/3} - -9^{3/2}\, dtheta

    With my limits from 0 to 2pi – wrong answer – I said “Ah the area under a sinusoidal – so 2 * the same integral from 0 to pi”

    Wrong again

    I then plot the polar plot of r = 3 sin(theta) and make a table and see that at pi/2 I get the first half of the “circle” polar plot.

    So I said, “Ah 4* the integral from 0 to pi/2 to get all the way back to 2pi” – see I had 2pi on the brain.

    Again – wrong answer, so I see that the plot completes the “circle” polar plot when the angle gets to pi.

    So when I multiply the integral by 2 and go from 0 to pi/2 I get it.

    Can you help clear the fog?

    When I see an integral of a sinusoid, my first instinct was to multiply the integral by 2 and then integrate from 0 to pi.

    I assume this is “Cartesian-thinking?”

    Next, I want to say when working with polar coordinates, that the angle should “sweep” all the way around from 0 to 2pi.

    I know the answer is associated with the plot of “r” – should one always generate a plot to see how “r” goes or is there a more straightforward way to set up the integral?

  2. jcsd
  3. Apr 8, 2009 #2
    Cone? r=3sin\theta

    Check that.
    x^2+y^2=3y on x-y plane
    Care to guess it is a cylinder? [Complete the square]

    So where is my bonus? Even a raise? Forget tenure....
  4. Apr 9, 2009 #3
    yes I meant cylinder

    the r = 3sin(theta) is correct.

    More than anything I'm wanting clarification and further explanation on setting up the limits with polar coordinates.

    Meaning - you see my first thought was to have theta go from 0 to 2pi.

    Then when I proceeded I had an integral involving sinusoidal functions so I said "double" the integral from 0 to pi.

    It wasn't until I plotted the graph and did a simple table (theta vs. r) that I saw that it's 2 times the integral from 0 to pi/2.

    did I miss an easier way to "see" this problem.

    What burned me was the integrating sin and it not being 2 times the integral from 0 to pi - like in Cartesian coordinates.

  5. Apr 11, 2009 #4
    Hello Sparky_,

    The way to tackle these problems is first of all to draw a clear picture. In doing so you have to think allready what the behaviour is of the equations of the shapes you have and thus you would have avoided the [itex]2 \pi[/itex] issue. Then try to see if there are any symmetry planes and if so split the volume up into smaller pieces of equal size so you only need to determine one part and mostly a simpler integral is a consequence. In this case I would say there are 4 pieces, 1 above the XY plane and 1 below, each of these latter ones cutted up in two smaller ones by the YZ plane. In order to set up the limits of the integrals try to find out what comes last. In this case it would be the [itex] \theta[/itex] dependency and the limits are now [itex]0\leq \theta \leq \pi/2[/itex] which you can see from your picture. After that it would be the r dependency having limits [itex]0\leq r \leq 3sin(\theta )[/itex]. The first integral to solve would be the one for the height z and the limits can be obtained from the equation of the sphere. You need to transform it into polar coordinates. [itex]x^2+y^2=r^2[/itex], so these are [itex]0 \leq z \leq \sqrt{9-r^2}[/itex]. The integral looks now as:

    [tex]V=4 \cdot \int_{0}^{\pi/2} \int_{0}^{3sin(\theta)} \int_{0}^{\sqrt{9-r^2}} dz \cdot r \cdot dr \cdot d\theta[/tex]

    The evaluation is up to you, I'm going to bed it's very late over here :-)

    Hope this helps,

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook