Chapter 14.8, Problem 24E

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

Chapter
Section

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
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 ) = ( 3 x + 2 y ) 2 2 y − x R: region bounded by the parallelogram with vertices (0, 0), (-2, 3), (2, 5), (4, 2)

To determine

To calculate: The volume of the provided solid region lying below the surface z=(3x+2y)22yx and above the plane region R.

Explanation

Given:

The provided function is f(x,y)=(3x+2y)22yâˆ’x.

The region R by parallelogram with vertices (2,5),(4,2),(0,0) and (âˆ’2,3).

Formula used:

Jacobians formula, âˆ‚(x,y)âˆ‚(u,v)=|âˆ‚xâˆ‚vâˆ‚xâˆ‚uâˆ‚yâˆ‚uâˆ‚yâˆ‚v|

And change of variables for double integrals is given as

âˆ¬Rf(x,y)dxdy=âˆ¬Sf(g(u,v),h(u,v))|âˆ‚(x,y)âˆ‚(u,v)|dudv

The slope-intercept form of equation of line y=mx+c, where â€˜mâ€™ is the slope of the line and m=y2âˆ’y1x2âˆ’x1.

Calculation:

The region R by parallelogram with vertices (2,5),(4,2),(0,0) and (âˆ’2,3). By using given coordinates, plot the parallelogram.

The slope of the straight line between the point (0,0) and (âˆ’2,3) is calculated as,

m=3âˆ’0âˆ’2âˆ’0=âˆ’32

The straight-line equation between the point (0,0) and (âˆ’2,3) is calculated as,

yâˆ’0=âˆ’32(xâˆ’0)2y=âˆ’3x3x+2y=0

The slope of straight line between the point (0,0) and (4,2) is calculated as,

m=2âˆ’04âˆ’0=12

The straight-line equation between the point (0,0) and (4,2) is calculated as,

yâˆ’0=12(xâˆ’0)2y=x2yâˆ’x=0

The slope of straight line between the point (2,5) and (4,2) is calculated as,

m=2âˆ’54âˆ’2=âˆ’32

The straight-line equation between the point (2,5) and (4,2) is calculated as,

yâˆ’2=âˆ’32(xâˆ’4)2yâˆ’4=âˆ’3x+123x+2y=16

The slope of straight line between the point (2,5) and (âˆ’2,3) is calculated as,

m=3âˆ’5âˆ’2âˆ’2=âˆ’2âˆ’4=12

The straight-line equation between the point (2,5) and (âˆ’2,3) is calculated as,

yâˆ’3=12(x+2)2yâˆ’6=x+22yâˆ’x=8

Assume that 2yâˆ’x=u and 3x+2y=v.

Subtracting these two equations, we get,

3x+x=vâˆ’ux=vâˆ’u4

Put x=vâˆ’u4 in the second equation,

2yâˆ’(vâˆ’u4)=u2y=v+3u4y=v+3u8

The line of the equation 2yâˆ’x=0 in new coordinate (u,v) is given as,

2yâˆ’x=0u=0

The line of the equation 3x+2y=0 in new coordinate (u,v) is given as,

3x+2y=0v=0

The line of the equation xâˆ’2y=âˆ’8 in new coordinate (u,v) is given as,

2yâˆ’x=8u=8

The line of the equation 3x+2y=0 in new coordinate (u,v) is given as,

3x+2y=16v=16

The straight line in a new rectangular co-ordinate is shown as below

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