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Thread: Help Wanted with Integral Word Problem

  1. #1
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    Help Wanted with Integral Word Problem

    Hello,

    I am not any good with these word problems.
    I am struggling with this:

    "Find the total mass of a circular region of radius 6 if the density (mass units per unit area) at a point in the region is the sqare of the distance from the point to the center of the circle. (Divide the region into circular shells, i.e., washers.)"


    I don't know where to begin, but a circle with radius 6. I can't even determine if this is supposed to be 2-D or 3-D.

    Can anyone please help me?
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  2. #2
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    Variable density disc

    Hello Jenberl
    Quote Originally Posted by Jenberl View Post
    Hello,

    I am not any good with these word problems.
    I am struggling with this:

    "Find the total mass of a circular region of radius 6 if the density (mass units per unit area) at a point in the region is the sqare of the distance from the point to the center of the circle. (Divide the region into circular shells, i.e., washers.)"


    I don't know where to begin, but a circle with radius 6. I can't even determine if this is supposed to be 2-D or 3-D.

    Can anyone please help me?
    It's a 2-D problem.

    Imagine a thin washer with centre at the centre of the disc, internal radius $\displaystyle x$ and external radius $\displaystyle x + \delta x$ (so its width is $\displaystyle \delta x$). Its internal circumference is $\displaystyle 2 \pi x$, and its area is therefore approximately $\displaystyle 2 \pi x \delta x$.

    The density at a distance $\displaystyle x$ from the centre is $\displaystyle x^2$ units of mass per unit of area. So the mass of this washer is

    $\displaystyle x^2 \times 2 \pi x \delta x = 2\pi x^3 \delta x$

    To get the mass of the whole disc you'll need to work out the integral of this expression (with $\displaystyle \delta x$ replaced by $\displaystyle dx$), with limits $\displaystyle x = 0$ to $\displaystyle x = 6$.

    Can you do it now?

    Grandad
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  3. #3
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    Smile

    Quote Originally Posted by Grandad View Post
    Hello Jenberl

    It's a 2-D problem.

    Imagine a thin washer with centre at the centre of the disc, internal radius $\displaystyle x$ and external radius $\displaystyle x + \delta x$ (so its width is $\displaystyle \delta x$). Its internal circumference is $\displaystyle 2 \pi x$, and its area is therefore approximately $\displaystyle 2 \pi x \delta x$.

    The density at a distance $\displaystyle x$ from the centre is $\displaystyle x^2$ units of mass per unit of area. So the mass of this washer is

    $\displaystyle x^2 \times 2 \pi x \delta x = 2\pi x^3 \delta x$

    To get the mass of the whole disc you'll need to work out the integral of this expression (with $\displaystyle \delta x$ replaced by $\displaystyle dx$), with limits $\displaystyle x = 0$ to $\displaystyle x = 6$.

    Can you do it now?

    Grandad

    Golly, thank you so much!
    How do you ever figure these problems out?
    I guess I don't have that foresight yet.
    Thank you again!
    Happy Holidays!

    Jen
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