   Chapter 14.8, Problem 26E

Chapter
Section
Textbook Problem

Finding Volume Using a Change of Variables In Exercises 23-30, use a change of variables to find the volume of the solid region lying below the surface z = f ( x , y ) and above the plane region R. f ( x , y ) = ( x + y ) 2 sin 2 ( x − y ) R: region bounded by the square with vertices ( π ,   0 ) ,   ( 3 π / 2 ,   π / 2 ) ,       ( π , π ) ,     ( π / 2 ,   π / 2 )

To determine

To calculate: The volume lying below the surface f(x,y)=(x+y)2sin(xy) and above the plane region R using change of variables.

Explanation

Given: The function f(x,y)=(x+y)2sin(xy)

R: Region bounded by the square with coordinates,

(x,y)=(π,0)(x,y)=(3π2,π2)(x,y)=(π,π)(x,y)=(π2,π2)

Formula used: Using Jacobian formula δ(x,y)δ(u,v)=|δxδuδxδvδyδuδyδv|

And, change of variables for double integrals

Rf(x,y)dxdy=Sf(g(u,v),h(u,v))|δ(x,y)δ(u,v)|dudv

We use the slope-intercept form of equation of line y=mx+c.

Where ‘m’ is the slope of the line and m=y2y1x2x1.

Calculation: Assuming, x=uv2 and y=u+v2

Givesus the value of u and v as,

u=x+y(Α)v=x+y(B)

Calculating the Jacobian as follows,

δ(x,y)δ(u,v)=|δxδuδxδvδyδuδyδv|δ(x,y)δ(u,v)=|12121212|δ(x,y)δ(u,v)=12

A graph is with the help of the given conditions using the slope-intercept form of equation of line y=mx+c.

Here ‘m’ is the slope of the line i.e. m=y2y1x2x1

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