Answer to Question #343961 in Calculus for Nina

Question

Find the mass of the lamina in the shape of the portion of the plane with equation 4x + 8y + z = 8 in the first octant if the area density at any point (x, y, z) on the plane is δ(x, y, z) = 6x + 12y + z g/cm^2

Answer 1

ANSWER The mass of the lamina is $96g$ .

EXPLANATION

The mass of the part of the plane $S$ is calculated using surface integral according to the formula $m=\iint_{S}\delta(x,y,z)dS$ , where $S= \left \{ \left ( x,y,z \right ): z=8-4x-8y,0\leq x\leq 2,\, 0\leq y\leq 1-\frac{x}{2} \right \}$ . $S$ has a one-to-one projection onto the domain $D$ in the $xy-$ plane:

$D=\left \{ \left ( x,y \right ): 0\leq x\leq 2,\, 0\leq y\leq 1-\frac{x}{2}\right \}$ . The surface area element on $S$ is given by $dS=\sqrt{1+z_{x}^{2}+z_{y}^{2}}\, dxdy=\sqrt{1+16+64}dxdy=9 dxdy$ . The surface integral of $\delta (x,y,z)$ over $S$ can be expressed as a double integral over the domain $D$

$amina \iint_{S} \delta (x,y,z)dS=\iint_{D} \delta (x,y,8-4x-8y)\cdot 9dxdy=\\=\iint_{D} (6x+12y+8-4x-8y)\cdot 9dxdy=18\iint_{D} ( x+2y+4 ) dxdy= \\=18\int_{0}^{2}\left ( \int_{0}^{1-\frac{x}{2}}\left ( x+2y+4 \right )dy \right )dx=18 \int_{0}^{2}\left \{\left [ \frac{\left ( x+2y+4 \right )^{2}}{4} \right ]_{y=0}^{1-\frac{x}{2}} \right \}dx=\\=18 \int_{0}^{2}\left ( 9-\frac{(x+4)^{2}}{4} \right )dx= 18 \left \{ 18-\left [ \frac{(x+4)^{3}}{12} \right ]_{0}^{2} \right \}= 18(18- \frac{6^{3}}{12}+\frac{4^{3}}{12})=18\cdot \frac{16}{3}=96$ .

So, $m=96$