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Math Help - Double Integrals

  1. #1
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    Double Integrals

    I have a couple of double integrals that I am having troubles with. I think for both of them it would be a good idea to use polar coordinates, but I am still having trouble setting them up/solving them.

    a) Double Integral (x dA), over D, where D is the region in the first quadrant that lies between the circles (x^2) + (y^2) = 4 and (x^2) + (y^2) = 2x.

    b) Find the volume of the given solid: inside the sphere (x^2) + (y^2) + (z^2) = 16 and outside the cylinder (x^2) + (y^2) = 4.

    Thanks for any help.
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  2. #2
    Member Mentia's Avatar
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    a)

    I am unsure what you mean by "x dA". I will proceed assuming there is no density function (nothing actually inside the integral besides dxdy). In this case,

    We know:

    r^2=x^2+y^2 and x = r \cos ( \theta )

    So the first circle is, in polar coordinates:
    r = 2

    The second circle is:
    r = 2 \cos (\theta)

    It helps to graph the two curves. You'll find that in the first quadrant you will be integrating from the smaller circle to the bigger circle:

    \int_{0}^{\frac {\pi}{2}} \int_{2 \cos (\theta)}^{2}rdrd\theta = \frac {\pi}{2}

    Here, theta from 0 to Pi/2 indicates the first quadrant, and r from 2 \cos (\theta) to 2 indicates you are going from the small circle to the big circle.

    This is easily checkable since we are dealing with perfect circles. If you look at the plot you see that you are subtracting half of a circle with radius 1 from 1/4 of a circle of radius 2. Thus: \frac{1}{4}\pi(2^2) - \frac{1}{2}\pi(1^2) = \frac{\pi}{2}

    Now, if you meant that you have a density function f(x) = x to integrate over, it changes things a bit:

    \int_{0}^{\frac {\pi}{2}} \int_{2 \cos (\theta)}^{2}rxdrd\theta = \int_{0}^{\frac {\pi}{2}} \int_{2 \cos (\theta)}^{2}r^2\cos (\theta)drd\theta = \frac{8}{3}-\frac{\pi}{2}
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  3. #3
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    Sorry, yes, I meant the density function is f(x) = x. Thanks for the help on part a).
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  4. #4
    Member Mentia's Avatar
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    b)

    Lets use cylindrical coordinates here.

    At what z do the cylinder and sphere meet?

    We have 16 - z^2 = 4 --> z = \pm  \sqrt[ ]{ 12}

    So now we know our limits on z. Once again,
    r^2=x^2+y^2

    So for the cylinder we have: r=2. For the sphere we have: r = \sqrt[]{16-z^2}.

    Then our integral becomes:

     \int_{0}^{2\pi }  \int_{ - \sqrt[ ]{ 12} }^{ \sqrt[ ]{12 }  }  \int_{2 }^{\sqrt[]{16-z^2}}  rdrdzd\theta = 32\,\sqrt[]{3}\pi
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  5. #5
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    For part b), what you described makes sense, but we haven't covered triple integrals or cylindrical coordinates in class yet, so I'm not completely following everything. Is it possible to do the question in polar coordinates, or is cylindrical the only method? Thanks for the help!
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  6. #6
    Member Mentia's Avatar
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    Okay so all cylindrical coordinates does it take polar coordinates into the z direction. It's like adding up a bunch of little discs of area given by polar coordinates and then multiplying by a little bit of height, dz, and integrating over that.

    Remember the equation for the volume of a cylinder:

    V = \pi r^2 h

    Notice that the volume is just the area of the base times the height. But what if the radius were non-constant? You can cut it up into small discs where the radius is constant and then add them all up.

    So if you look at a slight rearrangement of the integral from above:

    V = \int_{ - \sqrt[ ]{ 12} }^{ \sqrt[ ]{12 } } \int_{0}^{2\pi }  \int_{2 }^{\sqrt[]{16-z^2}} rdrd\theta dz = 32\,\sqrt[]{3}\pi<br />

    Notice that the inner two integrals are simply the polar coordinates area of the region between the cylinder and the sphere for a given z (radius changes as z changes, thats why the r limit has z in it).
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