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Text: Math 254 10/29 Concept Check 12.7 Triple Integrals

The formulas for the mass and center of mass for a 3D object are natural extensions of the formulas for a 2D object:

m =

âˆ p(x,y,z) dV

Myz âˆ x p(x,y,z) dV

Mxz âˆ y p(x,y,z) dV

Mxy âˆ 2 p(x,y,z) dV

Myz X = m

Mxz y = m

Mxy z = m

Suppose an object occupies the rectangular cube D = { (x,y,z) |-1<x <1,0 <y<4,-3<z <0} with a density function of p(x,y,z) = y. Find the total mass and center of mass:

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$3-10$ Find the mass and center of mass of the lamina that occupies

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m = âˆ«âˆ«âˆ« p(x, y, z) dV

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Find the mass and center of mass of the solid $E$ with the given density function $\rho .$

\begin{equation}

\begin{array}{l}{E \text { lies above the } x y \text { -plane and below the paraboloid }} \\ {z=1-x^{2}-y^{2} ; \quad \rho(x, y, z)=3}\end{array}

\end{equation}

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Find the mass and center of mass of the lamina that occupies the region and has the given density function D, which is bounded by Y = 1 X^2 and Y = 0; p(x, Y) = 7ky^2 (x,y) = 0.

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Transcript

So we have an tote. Tensity is to the z, 5 and 10. Defined is long to minus 1 to 1 y equal to 4 and z is minus 2 k, so we have to find x. Mass mass will be equal to the whole volume to x by d t so from here. We can write it as 0245. T y d d to d x is minus 1 to 1 and 2. As we can see, these are already determined that it has 5 steps 54 and the x minus 11 plus 1. So to get the answer to it to a ford 8 point now we have to find…

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