How to calculate the mass of an object in Vector Calculus?

 Let's consider a piece of wire and calculate its mass. We represent this piece of wire by a curve. The mass of an object is equal to its density by its length. The mass per unit length or linear density is a continuous function ρ(x, y, z). Let's consider a small piece Δs of the wire. Its mass is equal to its density multiplied by its length:  ρ(x, y, z)Δs. Let's add the masses of all the pieces and let each length Δs shrink to zero. The result is the integral ∫_C ρ(x, y, z) ds.

Example

Solution

The formula for the mass is

\[ m = \int_C \rho(x, y, z)\, ds \]

The formula of the scalar line integral is

\[ \int_C \rho(x, y, z)\, ds = \int_a^b \rho(\mathbf{r}(t))\, \|\mathbf{r}'(t)\|\, dt \]

Let's calculate \( \rho(\mathbf{r}(t)) \), \( \mathbf{r}'(t) \) and \( \|\mathbf{r}'(t)\| \)

Here \( \mathbf{r}(t) = \langle t,\ 2\cos t,\ 2\sin t\rangle \) for \( 0 \le t \le \frac{\pi}{2} \), so \( x = t \), \( y = 2\cos t \) and \( z = 2\sin t \).

\[ \rho(\mathbf{r}(t)) = e^{t} + (2\cos t)(2\sin t) = e^{t} + 4\sin t\cos t = e^{t} + 2\sin 2t \] \[ \mathbf{r}'(t) = \langle 1,\ -2\sin t,\ 2\cos t\rangle \] \[ \|\mathbf{r}'(t)\| = \sqrt{1 + 4\sin^2 t + 4\cos^2 t} = \sqrt{1 + 4} = \sqrt{5} \]

Let's calculate the integral

\[ \int_0^{\pi/2} \rho(\mathbf{r}(t))\, \|\mathbf{r}'(t)\|\, dt = \sqrt{5}\int_0^{\pi/2} \left(e^{t} + 2\sin 2t\right) dt \]

An antiderivative of \( e^{t} + 2\sin 2t \) is \( e^{t} - \cos 2t \), so:

\[ \sqrt{5}\Big[e^{t} - \cos 2t\Big]_0^{\pi/2} = \sqrt{5}\Big[(e^{\pi/2} - \cos \pi) - (e^{0} - \cos 0)\Big] = \sqrt{5}\left(e^{\pi/2} + 1\right) \]

Let's substitute the integral in the formula for \( m \)

\[ m = \int_C \rho(x, y, z)\, ds = \int_0^{\pi/2} \rho(\mathbf{r}(t))\, \|\mathbf{r}'(t)\|\, dt = \sqrt{5}\left(e^{\pi/2} + 1\right) \]

Conclusion. The mass of the spring is \( m = \sqrt{5}\left(e^{\pi/2} + 1\right) \approx 12.99 \) kg.