Questions: Suppose the density of a thin plate represented by the region (R) is (rho(r, theta)) (in units of mass per area). The mass of the plate is (iintR rho(r, theta) d A). Find the mass of the thin half annulus (R=(r, theta): 1 leq r leq 4,0 leq theta leq pi) with a density (rho(r, theta)=6+r sin theta). Find the mass. [ mathrmM=square ] (Type an exact answer.)

Solution Steps

To find the mass of the thin half annulus, we need to evaluate the double integral of the density function over the given region. The region is defined in polar coordinates, so we will integrate with respect to \( r \) from 1 to 4 and with respect to \( \theta \) from 0 to \( \pi \). The density function is given as \( \rho(r, \theta) = 6 + r \sin \theta \). The differential area element in polar coordinates is \( r \, dr \, d\theta \). Therefore, the mass is given by the integral:

\[ \mathrm{M} = \int_{0}^{\pi} \int_{1}^{4} (6 + r \sin \theta) \cdot r \, dr \, d\theta \]

To find the mass \( \mathrm{M} \) of the thin half annulus, we set up the double integral of the density function \( \rho(r, \theta) = 6 + r \sin \theta \) over the region defined by \( 1 \leq r \leq 4 \) and \( 0 \leq \theta \leq \pi \):

\[ \mathrm{M} = \int_{0}^{\pi} \int_{1}^{4} (6 + r \sin \theta) \cdot r \, dr \, d\theta \]

We first evaluate the inner integral with respect to \( r \):

\[ \int_{1}^{4} (6 + r \sin \theta) \cdot r \, dr = \int_{1}^{4} (6r + r^2 \sin \theta) \, dr \]

Calculating this gives:

\[ = \left[ 3r^2 + \frac{r^3}{3} \sin \theta \right]_{1}^{4} \]

Evaluating at the bounds:

\[ = \left( 3(4^2) + \frac{4^3}{3} \sin \theta \right) - \left( 3(1^2) + \frac{1^3}{3} \sin \theta \right) \]

\[ = \left( 48 + \frac{64}{3} \sin \theta \right) - \left( 3 + \frac{1}{3} \sin \theta \right) \]

\[ = 45 + \frac{63}{3} \sin \theta = 45 + 21 \sin \theta \]

Next, we evaluate the outer integral with respect to \( \theta \):

\[ \int_{0}^{\pi} (45 + 21 \sin \theta) \, d\theta \]

Calculating this gives:

\[ = \left[ 45\theta - 21 \cos \theta \right]_{0}^{\pi} \]

Evaluating at the bounds:

\[ = \left( 45\pi - 21(-1) \right) - \left( 45(0) - 21(1) \right) \]

\[ = 45\pi + 21 + 21 = 45\pi + 42 \]

Final Answer