Find the center of mass of the solid in Example 5.

Question

Accepted Answer

We have already found that [latex]\mu=\frac{47}{84}[/latex]. We next calculate the moments.

[latex]M_{y z}=\int_{0}^{1} \int_{x^{2}}^{x} \int_{x-y}^{x+y} x(x+2 y+4 z) d z d y d x[/latex]

[latex]\begin{aligned}&=\int_{0}^{1} \int_{x^{2}}^{x}\left\{\left.\left(\left(x^{2}+2 x y ight) z+2 x z^{2} ight) ight|_{x-y} ^{x+y} ight\} d y d x \&=\int_{0}^{1} \int_{x^{2}}^{x}\left(10 x^{2} y+4 x y^{2} ight) d y d x=\int_{0}^{1}\left\{\left.\left(5 x^{2} y^{2}+\frac{4 x y^{3}}{3} ight) ight|_{x^{2}} ^{x} ight\} d x \&=\int_{0}^{1}\left(5 x^{4}-5 x^{6}+\frac{4}{3} x^{4}-\frac{4}{3} x^{7} ight) d x=1-\frac{5}{7}+\frac{4}{15}-\frac{1}{6}=\frac{27}{70} \M_{x z} &=\int_{0}^{1} \int_{x^{2}}^{x} \int_{x-y}^{x+y} y(x+2 y+4 z) d z d y d x \&=\int_{0}^{1} \int_{x^{2}}^{x}\left\{\left.\left[\left(x y+2 y^{2} ight) z+2 y z^{2} ight] ight|_{x-y} ^{x+y} ight\} d y d x \&=\int_{0}^{1} \int_{x^{2}}^{z}\left(10 x y^{2}+4 y^{3} ight) d y d x=\int_{0}^{1}\left\{\left.\left(\frac{10}{3} x y^{3}+y^{4} ight) ight|_{x^{2}} ^{x} ight\} d x \&=\int_{0}^{1}\left(\frac{10}{3} x^{4}-\frac{10}{3} x^{7}+x^{4}-x^{8} ight) d x=\frac{2}{3}-\frac{5}{12}+\frac{1}{5}-\frac{1}{9}=\frac{61}{180}\end{aligned}[/latex]

[latex]\begin{aligned}M_{x y} &=\int_{0}^{1} \int_{x^{2}}^{x} \int_{x-y}^{x+y} z(x+2 y+4 z) d z d y d x \&=\int_{0}^{1} \int_{x^{2}}^{x}\left\{\left.\left[(x+2 y) \frac{z^{2}}{2}+\frac{4 z^{3}}{3} ight] ight|_{x-y} ^{x+y} ight\} d y d z \&=\int_{0}^{1} \int_{x^{2}}^{x}\left[(x+2 y)(2 x y)+\frac{4}{3}\left(6 x^{2} y+2 y^{3} ight) ight] d y d x \&=\int_{0}^{1} \int_{x^{2}}^{x}\left(10 x^{2} y+4 x y^{2}+\frac{8}{3} y^{3} ight) d y d x \&=\int_{0}^{1}\left\{\left.\left(5 x^{2} y^{2}+\frac{4}{3} x y^{3}+\frac{2}{3} y^{4} ight) ight|_{x^{2}} ^{x} ight\} d x \&=\int_{0}^{1}\left(5 x^{4}-5 x^{6}+\frac{4}{3} x^{4}-\frac{4}{3} x^{7}+\frac{2}{3} x^{4}-\frac{2}{3} x^{8} ight) d x \&=\int_{0}^{1}\left(7 x^{4}-5 x^{6}-\frac{4}{3} x^{7}-\frac{2}{3} x^{8} ight) d x=\frac{7}{5}-\frac{5}{7}-\frac{1}{6}-\frac{2}{27}=\frac{841}{1890}\end{aligned}[/latex]

Thus

[latex]\begin{aligned}(\bar{x}, \bar{y}, \bar{z}) &=\left(\frac{M_{y z}}{\mu}, \frac{M_{x z}}{\mu}, \frac{M_{x y}}{\mu} ight)=\left(\frac{27 / 70}{47 / 84}, \frac{61 / 180}{47 / 84}, \frac{841 / 1890}{47 / 84} ight) \&=\left(\frac{162}{235}, \frac{427}{705}, \frac{1682}{2115} ight) \approx(0.689,0.606,0.795)\end{aligned}[/latex]

Here distances are measured in meters.

Question 5.6.6: Find the center of mass of the solid in Example 5.

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