Help Wanted with Integral Word Problem

**Jenberl** · December 23rd 2008, 05:46 AM

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?

**Grandad** · December 23rd 2008, 06:01 AM

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.

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 $x$ and external radius $x + \delta x$ (so its width is $\delta x$ ). Its internal circumference is $2 \pi x$ , and its area is therefore approximately $2 \pi x \delta x$ .

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

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

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

Can you do it now?

Grandad

**Jenberl** · December 23rd 2008, 06:25 AM

Originally Posted by Grandad

Hello Jenberl

It's a 2-D problem.

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

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

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

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

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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