15.6 triple integrals: problem 6 (1 point) find the center mass of the solid bounded by planes x + y + z =…

Explanation:

Step1: Find the mass (M)

The limits of integration for the solid are (0\leq z\leq 1 - x - y), (0\leq y\leq 1 - x), (0\leq x\leq 1). The mass (M) is given by the triple - integral (M=\int_{0}^{1}\int_{0}^{1 - x}\int_{0}^{1 - x - y}14\sqrt{z}dzdydx). First, integrate with respect to (z): (\int_{0}^{1 - x - y}14\sqrt{z}dz=14\times\frac{2}{3}z^{\frac{3}{2}}\big|{0}^{1 - x - y}=\frac{28}{3}(1 - x - y)^{\frac{3}{2}}). Then integrate with respect to (y): (\int{0}^{1 - x}\frac{28}{3}(1 - x - y)^{\frac{3}{2}}dy=-\frac{28}{3}\times\frac{2}{5}(1 - x - y)^{\frac{5}{2}}\big|{0}^{1 - x}=\frac{56}{15}(1 - x)^{\frac{5}{2}}). Finally, integrate with respect to (x): (\int{0}^{1}\frac{56}{15}(1 - x)^{\frac{5}{2}}dx=-\frac{56}{15}\times\frac{2}{7}(1 - x)^{\frac{7}{2}}\big|_{0}^{1}=\frac{16}{15}).

Step2: Find (M_{yz}) (moment about the (yz) - plane)

(M_{yz}=\int_{0}^{1}\int_{0}^{1 - x}\int_{0}^{1 - x - y}x\cdot14\sqrt{z}dzdydx). Integrating with respect to (z) first gives (\int_{0}^{1 - x - y}x\cdot14\sqrt{z}dz=x\cdot\frac{28}{3}(1 - x - y)^{\frac{3}{2}}). Integrating with respect to (y) gives (\int_{0}^{1 - x}x\cdot\frac{28}{3}(1 - x - y)^{\frac{3}{2}}dy=-x\cdot\frac{56}{15}(1 - x)^{\frac{5}{2}}). Integrating with respect to (x): (\int_{0}^{1}- \frac{56}{15}x(1 - x)^{\frac{5}{2}}dx). Let (u = 1 - x), then (x=1 - u) and (dx=-du). The integral becomes (\frac{56}{15}\int_{0}^{1}(1 - u)u^{\frac{5}{2}}du=\frac{56}{15}(\int_{0}^{1}u^{\frac{5}{2}}du-\int_{0}^{1}u^{\frac{7}{2}}du)=\frac{56}{15}(\frac{2}{7}-\frac{2}{9})=\frac{128}{135}).

Step3: Find (M_{xz}) (moment about the (xz) - plane)

By symmetry (since the region and the density function are symmetric with respect to (x) and (y)), (M_{xz}=M_{yz}=\frac{128}{135}).

Step4: Find (M_{xy}) (moment about the (xy) - plane)

(M_{xy}=\int_{0}^{1}\int_{0}^{1 - x}\int_{0}^{1 - x - y}z\cdot14\sqrt{z}dzdydx=\int_{0}^{1}\int_{0}^{1 - x}\int_{0}^{1 - x - y}14z^{\frac{3}{2}}dzdydx). Integrating with respect to (z) first: (\int_{0}^{1 - x - y}14z^{\frac{3}{2}}dz=14\times\frac{2}{5}z^{\frac{5}{2}}\big|{0}^{1 - x - y}=\frac{28}{5}(1 - x - y)^{\frac{5}{2}}). Integrating with respect to (y): (\int{0}^{1 - x}\frac{28}{5}(1 - x - y)^{\frac{5}{2}}dy=-\frac{28}{5}\times\frac{2}{7}(1 - x - y)^{\frac{7}{2}}\big|{0}^{1 - x}=\frac{8}{5}(1 - x)^{\frac{7}{2}}). Integrating with respect to (x): (\int{0}^{1}\frac{8}{5}(1 - x)^{\frac{7}{2}}dx=-\frac{8}{5}\times\frac{2}{9}(1 - x)^{\frac{9}{2}}\big|_{0}^{1}=\frac{16}{45}).

Step5: Calculate the center - of - mass coordinates

(x_{CM}=\frac{M_{yz}}{M}=\frac{\frac{128}{135}}{\frac{16}{15}}=\frac{8}{9}), (y_{CM}=\frac{M_{xz}}{M}=\frac{8}{9}), (z_{CM}=\frac{M_{xy}}{M}=\frac{\frac{16}{45}}{\frac{16}{15}}=\frac{1}{3}).

Answer:

((\frac{8}{9},\frac{8}{9},\frac{1}{3}))