Question

find the moment about the x-axis of a wire of constant density that lies along the curve y=2(squarerootx) from x=0 to x=3

Answer 1

Here's a start. (I'm assuming by "moment" you mean moment of inertia). Let the mass-per-length density of the wire be \(\lambda\). Thus, if \(ds\) is an infinitesimal arclength, we have \(\lambda = \frac{dm}{ds}\). Also, let \(r\) be the distance of the wire element to the axis of rotation, which in this case is \(r=y\).\[I=\int\limits_\text{wire} r^2 dm = \int\limits_\text{wire} y^2 \left( \lambda\ ds\right)\]Can you take it from there? (If you need more help, consider that \(ds^2=dy^2+dx^2\).)

Answer 2

i honestly have no idea what to do at all.

Answer 3

so is it 4x(1/squareroot x)

Answer 4

\(y^2=4x\) is correct, though be careful with \(ds\).\[ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx\] should help. This is the same technique one uses to calculate arclength of a function.

Answer 5

4x\[\sqrt{1+1/\sqrt{x}}\]

Answer 6

that should be 4x times the radical

Answer 7

Don't forget to square \(\frac{dy}{dx}\), but yes. Now integrate.\[I=\int\limits_\text{wire} 4\lambda x\sqrt{1+\frac{1}{x}}\ dx\]WolframAlpha says that this is a very tricky integral...let me double-check this for accuracy.

Answer 8

I don't see any errors in my work... I'd just evaluate the integral between the bounds by WolframAlpha or using a graphing calculator.

Answer 9

ok thank you so much

Answer 10

You're welcome.