Center of Mass

**Spudwad** · February 21st 2010, 06:32 PM

Here is the question:

A solid is formed by rotating the region bounded by the curve

and the

-axis between

and

, around the

-axis. The volume of this solid is

. Assuming the solid has constant density

, find

and

.

To find the moment, I did:

$\int_0^1 x dx$ =

$\frac {1} {2}$

As density was constant, I divided the moment by the volume and found

to be 1/2.

And I am not sure that is right and so I haven't even attempted to find

. All help is much appreciated. Thanks.

**HallsofIvy** · February 22nd 2010, 01:01 AM

The volume, when y= y(x) is rotated around the x- axis is (by the disk method) $\pi \int_{x_0}^{x_1} y^2(x)dx$ .

In this case, that would be $\pi \int_0^1 e^{-7x} dx$ which gives the volume you quote.

But the moment is NOT just that volume times $\int x dx$ . It is, rather, that same integral with another x in the integral: $\pi \int_0^1 xe^{-7x}dx$ . Use "integration by parts".

Since this is rotated around the x-axis, it is symmetric with respect to both y and z and it immediately follows that $\overline{y}= 0$ and $\overline{z}= 0$ .

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