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Chapter 5: Q46P (page 248)

Find the mass of a cube of side 2 if the density is proportional to the square of the distance from the center of the cube.

Short Answer

Expert verified

The required Mass of a cube is $8K$ .

Step by step solution

01

Definition of Mass

A physical body's mass is the amount of matter it contains. It's also a measure of inertia, or the resistance to acceleration when a net force is applied.

02

Finding the Mass of a cube

The origin should be in the center of the cube. We may compute the mass in one octant and then double the result by 8 because the cube's density is symmetrical in x, y and z .

Now, Find the Mass,

$M = 8 K \int_{0}^{1} (x^{2} + y^{2} + z^{2}) d x \int_{0}^{1} d y \int_{0}^{1} d z = {8 K \int_{0}^{1} d x \int_{0}^{1} d y (x^{2} z + y^{2} z + \frac{1}{3} z^{3})|}_{0}^{1} = 8 K \int_{0}^{1} d x \int_{0}^{1} d y (x^{2} + y^{2} + \frac{1}{3}) = {8 K \int_{0}^{1} d x (x^{2} y + \frac{1}{3} y^{3} + \frac{1}{3} y)|}_{0}^{1} = 8 K \int_{0}^{1} d x (x^{2} + \frac{1}{3} + \frac{1}{3})$

The final value is obtained using further calculation.

$M = 8 K \int_{0}^{1} (x^{2} + \frac{2}{3}) d x = {8 K [\frac{x^{3}}{3} + \frac{2}{3} x]|}_{0}^{1} = 8 K (\frac{1}{3} + \frac{2}{3}) = 8 K$

M = 8 K \int_{0}^{1} (x^{2} + \frac{2}{3}) d x = {8 K [\frac{x^{3}}{3} + \frac{2}{3} x]|}_{0}^{1} = 8 K (\frac{1}{3} + \frac{2}{3}) = 8 K

Therefore, the Mass of a cube is $8 K$ .

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Most popular questions from this chapter

(a) Write a triple integral in cylindrical coordinates for the volume of the part of a ball between two parallel planes which intersect the ball.

(b) Evaluate the integral in (a). Warning hint: Do the r and $θ$ integrals first.

(c) Find the centroid of this volume.

Find theJacobian $\partial (x, y) / \partial (u, y)$ of the given transformations from variables x,yto variables u,v:

$_{y = a s i n h u s i n v,}^{x = a c o s h u c o s v,} (u and v are called elliptic cylinder coordinates)$

The volume inside a sphere of radius ris $V = \frac{4}{3} {ττr}^{3}$ . Then $dV = 4 {ττr}^{2} dr = Adr$ whereis the area of the sphere. What is the geometrical meaning of the fact that the derivative of the volume is the area? Could you use this fact to find the volume formula given the area formula?

$\int_{x = 0}^{4} \int_{y = 0}^{\sqrt{x}} \sqrt[y]{x} dydx$

Find the area of the part of the cone $2 z^{2} = x^{2} + y^{2}$ in the first octant cut out by the planes y = 0 and $y = \frac{x}{\sqrt{3}}$ ,and the cylinder

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