# Thread: Help Wanted with Integral Word Problem

1. ## 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.

2. ## Variable density disc

Hello Jenberl
Originally Posted by Jenberl
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.

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?

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?