   Chapter 14.4, Problem 5E

Chapter
Section
Textbook Problem

Finding the Mass of a Lamina In Exercises 3-6, find the mass of the lamina described by the inequalities, given that its density is ρ ( x , y ) = x y . 0 ≤ x ≤ 1 ,       0 ≤ y ≤ 1 − x 2

To determine

To calculate: The mass of the lamina described by the inequalities given that its density is

ρ(x,y)=xy, 0x1, 0y1x2.

Explanation

Given:

The provided density of lamina is: ρ(x,y)=xy, 0x1, 0y1x2.

Formula used:

Mass of planar lamina is:

m=Rρ(x,y)dA

The integral of rndr=rn+1n+1+c.

The relation from trigonometric identity,

sin2θ=2sinθcosθ

Calculation:

Consider the density function ρ(x,y)=xy, 0x1, 0y1x2.

Here the limit of y is, 0y1x2 and that of x 0x1.

The given limits for the integral has square root function, so it is easier to solve this problem by

the use of polar coordinates.

First convert the double integral into polar coordinates as:

x=rcosθy=rsinθdxdy=rdrdθ

The limits of the given double integral are:

0x10y1x2

As y varies from 0 to 1x2.

As the equation of circle is x2+y2=r2 and compare y=1x2 with the equation of circle

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