Q. 16 - Calculus

Question

In Exercises $15-17$, evaluate the three integrals that can be used to find the moments of inertia for the pyramid $Q$ described in Example $5$ and then use those values to find the radii of gyration about the coordinate axes. Recall that the mass of $Q$ is $\frac{1}{6}k$.

Show that $I_y={\int }_0^1{\int }_0^{-x+1}{\int }_0^{-x-y+1}k\left(y^2+z^2\right)dzdydx=\frac{1}{30}k$. Use your answer to show that $R_y=\frac{\sqrt{5}}{5}$.

Step-by-Step Solution

Answer

The given equation ${\int }_0^1{\int }_0^{-x+1}{\int }_0^{-x-y+1}k\left(y^2+z^2\right)dzdydx=\frac{1}{30}k$ is verified and also $R_y=\frac{\sqrt{5}}{5}$ is showed.

Step 1 . Given information

$$\begin{gathered} m=\frac{k}{6} \\ {\int }_0^1{\int }_0^{-x+1}{\int }_0^{-x-y+1}k\left(y^2+z^2\right)dzdydx=\frac{1}{30}k \\ {R}_y=\frac{\sqrt{5}}{5} \end{gathered}$$

Step 2 . Consider the triple integral,

$I_y={\int }_0^1{\int }_0^{-x+1}{\int }_0^{-x-y+1}k\left(y^2+z^2\right)dzdydx$.

Evaluate the value of the given triple integral in the following way:

$$\begin{gathered} {\int }_0^1{\int }_0^{-x+1}{\int }_0^{-x-y+1}k\left(y^2+z^2\right)dzdydx={\int }_0^1{\int }_0^{-x+1}\left({\int }_0^{-x-y+1}k\left(y^2+z^2\right)dz\right)dydx \\ =k{\int }_0^1{\int }_0^{-x+1}\left(y^2{\int }_0^{-x-y+1} dz+{\int }_0^{-x-y+1}{z}^2.dz\right)dydx \\ =k{\int }_0^1{\int }_0^{-x+1} \left(y^2{\left[z\right]}_0^{-x-y+1}+{\left[\frac{z^3}{3}\right]}_0^{-x-y+1}\right)dydx \\ =k{\int }_0^1{\int }_0^{-x+1}\left(y^2\left(1-x-y\right)+\frac{1}{3}\left({\left(-x-y+1\right)}^3-0\right)\right)dydx \\ =k{\int }_0^1{\int }_0^{-x+1}\left(y^2\left(1-x-y\right)+\frac{1}{3}{\left(-x-y+1\right)}^3\right)dydx \end{gathered}$$

Step 3 . Further, simplify the above right hand side integral as follows:

The formula for ${\left(a-b-c\right)}^2$ is as follows:

${\left(a-b-c\right)}^2=a^2+b^2+c^2-2ab+2bc-2ca$.

Use the above formula to expand the term in the following step.

$$\begin{gathered} {\int }_0^1{\int }_0^{-x+1}{\int }_0^{-x-y+1}k\left(y^2+z^2\right)dzdydx=k{\int }_0^1{\int }_0^{-x+1}\left(\left(1-x-y\right)\left(y^2+\frac{{\left(1-x-y\right)}^2}{3}\right)\right)dydx \\ =k{\int }_0^1{\int }_0^{-x+1}\left(\left(1-x-y\right)\left(\frac{3y^2+1+x^2+y^2-2x+2yx-2y}{3}\right)\right)dydx \end{gathered}$$

Step 4 . First take the inner terms and simplify it.

$$\begin{gathered} \left(1-x-y\right)\left(\frac{3y^2+1+x^2+y^2-2x+2yx-2y}{3}\right)=\frac{1}{3}\left[\left(1-x-y\right)\left(4y^2+1+x^2-2x+2yx-2y\right)\right] \\ =\frac{1}{3}\left[4y^2-4x{y}^2-4y^3+1-x-y+x^2-x^3-x^{2}y-2x+2x^2+2xy+2xy-2x^{2}y-2x{y}^2-2y+2xy+2y^2\right] \\ =\frac{1}{3}\left[1-3x-3y+3x^2+6y^2-x^3-4y^3+6xy-3x^{2}y-6x{y}^2\right] \end{gathered}$$

Step 5 . Now substitute the known values in the integral.

$$\begin{gathered} {\int }_0^1{\int }_0^{-x+1}{\int }_0^{-x-y+1}k\left(y^2+z^2\right)dzdydx=\frac{k}{3}{\int }_0^1{\int }_0^{-x+1}\left[1-3x-3y+3x^2+6y^2-x^3-4y^3+6xy-3x^{2}y-6x{y}^2\right]dydx \\ =\frac{k}{3}{\int }_0^1\left[{\left[y\right]}_0^{1-x}-3x{\left[y\right]}_0^{1-x}-3{\left[\frac{y^2}{2}\right]}_0^{1-x}+3x^2{\left[y\right]}_0^{1-x}+6{\left[\frac{y^3}{3}\right]}_0^{1-x}-x^3{\left[y\right]}_0^{1-x}-4{\left[\frac{y^4}{4}\right]}_0^{1-x}+6x{\left[\frac{y^2}{2}\right]}_0^{1-x}-3x^2{\left[\frac{y^2}{2}\right]}_0^{1-x}-6x{\left[\frac{y^3}{3}\right]}_0^{1-x}\right]dx \\ =\frac{k}{3}{\int }_0^1\left[\left(1-x\right)-3x\left(1-x\right)-\frac{3}{2}{\left(1-x\right)}^2+3x^2\left(1-x\right)+2{\left(1-x\right)}^3-x^3\left(1-x\right)-{\left(1-x\right)}^4+3x{\left(1-x\right)}^2-\frac{3}{2}{x}^2{\left(1-x\right)}^2-2x{\left(1-x\right)}^3\right]dx \end{gathered}$$

Therefore, ${\int }_0^1{\int }_0^{-x+1}{\int }_0^{-x-y+1}k\left(y^2+z^2\right)dzdydx=\frac{1}{30}k$ is showed.

Step 6 . Given value for mass is, m = k 6 .

By definition, $R_y=\sqrt{\frac{I_y}{m}}$.

$$\begin{gathered} R_y=\sqrt{\frac{{\displaystyle \frac{k}{30}}}{{\displaystyle \frac{k}{6}}}} \\ =\sqrt{\frac{6}{30}} \\ =\sqrt{\frac{1}{5}} \\ =\frac{\sqrt{5}}{5} \end{gathered}$$

Hence, $R_y=\frac{\sqrt{5}}{5}$ is showed.