Question

A lamina covering the quarter disk ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{\le }\mathbf{4}\mathbf{,}\mathbf{x}\mathbf{>}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{>}\mathbf{0}$ has (area) density . Find the mass of the lamina.

Short answer

The mass of the lamina is,$\frac{16}{3}·$

Step-by-step solution

Definition of double integral and mass formula

The double integral of $\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$ over the areaA in the$\left(x,y\right)$ plane as the limit of this sum, and we write it as$\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathbf{f}\left(x,y\right)\mathbf{dxdy}$

Using the double integral the mass of the lamina:$\mathbf{M}\mathbf{=}\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathbf{\rho dA}\mathbf{.}$

Calculation of the mass of the lamina

To get the mass of the lamina, the area of integration is the quarter-circle with a radius$\mathrm{R}=2$.

The integrals:

$$\begin{gathered} \mathrm{M}=\int {\int }_{\mathrm{A}}\mathrm{\rho dA} \\ =\int {\int }_{\mathrm{A}}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{dxdy} \end{gathered}$$

Further calculation of the mass of the lamina

Change into the polar coordinate system,

$$\begin{gathered} \mathrm{M}={\int }_0^2{\int }_0^{\mathrm{\pi }/2}\mathrm{r}\left(\cos \mathrm{\theta }+\sin \mathrm{\theta }\right)\mathrm{rdrd\theta } \\ ={\int }_0^2{\mathrm{r}}^2\mathrm{dr}{\int }_0^{\mathrm{\pi }/2}\left(\cos \mathrm{\theta }+\sin \mathrm{\theta }\right)\mathrm{d\theta } \\ ={\overline{)\frac{8}{3}\left(\mathrm{sin\theta }-\mathrm{cos\theta }\right)}}_0^{\mathrm{\pi }/2} \\ =\frac{8}{3}\left(1-0-0+1\right) \\ =\frac{16}{3} \end{gathered}$$

Therefore, the mass of the lamina is, $\frac{16}{3}$.