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Chapter 5: Q50P (page 249)

Find the mass of the solid in Problem 48 if the density is z.

Short Answer

Expert verified

The required Mass of a solid is $182.04$ .

Step by step solution

01

Definition of Mass of a solid

A solid mass is the amount of a substance contained by a body, measured in Kilograms.

02

Finding the Mass of a solid

Consider the given density z,

Now, find the mass:

$M = {鈭�}_{0}^{4} dx {鈭�}_{0}^{x} dy {鈭�}_{0}^{x^{2} {-y}^{2}} zdz = {{鈭�}_{0}^{4} dx {鈭�}_{0}^{x} dy \frac{1}{2} z^{2}|}_{0}^{2} {-y}^{2} = \frac{1}{2} {鈭�}_{0}^{4} dx {鈭�}_{0}^{x} dy (x^{4} {- 2x}^{2} y^{2} {+y}^{4}) = {\frac{1}{2} {鈭�}_{0}^{4} dx (x^{4} y - \frac{2}{3} x^{2} y^{3} + \frac{y^{5}}{5})|}_{0}^{x}$

The final value is obtained using further calculation 鈥�

$M = \frac{1}{2} {鈭�}_{0}^{4} (x^{5} - \frac{2}{3} x^{5} + \frac{x^{5}}{5}) dx = \frac{4}{15} {鈭�}_{0}^{4} x^{5} dx = \frac{4}{90} x^{6} {鈭�}_{0}^{4}$

$M = \frac{8192}{45} = 182.04$

Hence, the Mass of a solid is $182.04$ .

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Most popular questions from this chapter

(a) Revolve the curve $y = x^{1}$ , from $x = 1 to x = 鈭�$ , about the x axis to create a surface and a volume. Write integrals for the surface area and the volume. Find the volume, and show that the surface area is infinite. Hint: The surface area integral is not easy to evaluate, but you can easily show that it is greater than ${鈭�}_{1}^{鈭�} x^{- 1} dx$ which you can evaluate.

(b) The following question is a challenge to your ability to fit together your mathematical calculations and physical facts: In (a) you found a finite volume and an infinite area. Suppose you fill the finite volume with a finite amount of paint and then pour off the excess leaving what sticks to the surface. Apparently, you have painted an infinite area with a finite amount of paint! What is wrong? (Compare Problem 15.31c of Chapter 1.)

Find the surface area cut from the cone $2 x^{2} + 2 y^{2} = 5 z^{2}, z > 0$ by the cylinder $x^{2} + y^{2} = 2 y$

Let the solid in Problem 7 have density $= c o s 胃$ .

Show that then $I_{z} = \frac{3}{10} M a^{2} s i n^{2} 伪$ .

A uniform chain hangs in the shape of the catenarybetween $x = - 1$ and $x = 1$ . Find

(a) its length,

(b)role="math" localid="1659154616792" $y$ .

Express the integral $I = {鈭�}_{0}^{1} d x {鈭�}_{0}^{\sqrt{1 - x^{2}}} e^{- x^{2} - y^{2}} d y$ as an integral in polar coordinates $(r, 胃)$ and so evaluate it.

See all solutions

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