# Thread:

1. ## Center of Mass

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.

2. 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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