Question

A triangular lamina is bounded by the coordinate axes and the line $\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{6}$. Find its mass if its density at each point P is proportional to the square of the distance from the origin to P.

Short answer

The mass of the triangular lamina is,$216\mathrm{K}$.

Step-by-step solution

Definition of double integral and mass formula

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

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

Diagram of the triangular lamina

The figure of the triangular lamina is bounded by the coordinate axes and the line $\mathrm{x}+\mathrm{y}=6$,

Calculation of the mass of the triangular lamina

The mass of the lamina is given by,

$$\begin{gathered} \mathrm{M}=\int {\int }_{\mathrm{A}}\mathrm{\rho dA} \\ ={\int }_0^6\left[{\int }_0^{6-\mathrm{x}}\mathrm{K}\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)\mathrm{dy}\right]\mathrm{dx} \\ =\mathrm{K}{\int }_0^6{\overline{)\left({\mathrm{x}}^2\mathrm{y}+\frac{1}{3}{\mathrm{y}}^3\right)}}_0^{6-\mathrm{x}}\mathrm{dx} \\ =\mathrm{K}{\int }_0^6\left({\mathrm{x}}^2\left(6-\mathrm{x}\right)+\frac{1}{3}{\left(6-\mathrm{x}\right)}^3\right)\mathrm{dx} \\ =\mathrm{K}{\overline{)\left[2{\mathrm{x}}^3-\frac{1}{4}{\mathrm{x}}^4\right]}}_0^6-{\int }_0^6{\left(6-\mathrm{x}\right)}^3\mathrm{d}\left(6-\mathrm{x}\right) \\ =\mathrm{K}\left(432-324\right)-\frac{\mathrm{K}}{12}{\overline{)\left[6-\mathrm{x}\right]}}_0^6 \\ =108\mathrm{K}+108\mathrm{K} \\ =216\mathrm{K} \end{gathered}$$

where K is the proportionality constant.

Therefore, the mass of the triangular lamina is,$216\mathrm{K}$.