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Chapter 5: Q44P (page 248)

Find the mass of the solid in Problem 42 if the density is proportional to y.

Short Answer

Expert verified

The required Mass of the solid is $\frac{7}{3} K$ .

Step by step solution

01

Definition of Mass

Double integralof $f(x,y)$ over the areaA in theplane $(x,y)$ as the limit of this sum, and write it as $\iint_{A} f(x,y)dxdy$ .

02

Finding the Mass

The given planes are, $z = 2 x + 3 y + 6 a$ and $z = 2 x + 7 y + 8$ .

Consider the density to be $ρ = K y$ and find the value of the mass.

$M = \int_{V} ρ d V = K \int_{0}^{1} d x \int_{0}^{1} y d y \int_{2 x + 3 y + 6}^{2 x + 7 y + 8} d z = K \int_{0}^{1} d x \int_{0}^{1} y d y (4 y + 2)$

The final value is obtained use it for further calculation.

$M = K \int_{0}^{1} d x \int_{0}^{1} d y (4 y^{2} + 2 y) = {K [\frac{4}{3} y^{3} + y^{2}]|}_{0}^{1} = \frac{7}{3} K$

Therefore, the value of the Mass is $\frac{7}{3} K$ .

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Above the square with vertices at, (0,0), (2,0),(0,2) and (2,2) and under the plane z = 8-x+y.

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